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25 tháng 4 2020
Bài 3:
Vì MN//BC, áp dụng định lí Talet, ta có:
\(\frac{AM}{AB}=\frac{AN}{AC}\Leftrightarrow\frac{3}{9}=\frac{4}{AC}\\ \Rightarrow AC=\frac{4\cdot9}{3}=12\\ \Rightarrow NC=AC-AN=12-4=8\)
Xét \(\Delta ABC,\widehat{A}=90^0\) , áp dụng định lí Pytago, ta có:
\(BC^2=AB^2+AC^2=9^2+12^2=225\\ \Rightarrow BC=\sqrt{225}=15\)
Tương tự, ta lại có MN//BC, nên:
\(\frac{AN}{AC}=\frac{MN}{BC}\Leftrightarrow\frac{4}{12}=\frac{MN}{15}\\ \Rightarrow MN=\frac{15.4}{12}=5\)
Xét \(\Delta ABN,\widehat{A}=90^0\) , áp dụng định lí Pytago, ta có:
\(BN^2=AB^2+AN^2=9^2+4^2=97\\ \Rightarrow BN=\sqrt{97}\approx9.8\)
Vậy \(NC=8\\ BC=15\\ MN=5\\ BN=9.8\)
25 tháng 4 2020
Bài 2: (Hình tự vẽ nha)
Vì \(MN\perp AB,AC\perp AB\) nên MN//AC.
Vì MN//AC (cmt), áp dung định lí Talet, ta có:
\(\frac{MB}{AB}=\frac{NB}{BC}\Leftrightarrow\frac{3}{AB}=\frac{5}{7}\\ \Rightarrow AB=\frac{3\cdot7}{5}=4.2\)
Xét \(\Delta ABC\), \(\widehat{A}=90^0\) , áp dụng định lí Pytago, ta có:
\(BC^2=AB^2+AC^2\\ \Rightarrow AC^2=BC^2-AB^2\\ \Leftrightarrow AC^2=7^2-4.2^2=31.36\\ \Rightarrow AC=\sqrt{31.36}=5.6\)
Chu vi của \(\Delta ABC\) là:
\(AB+AC+BC=4.2+7+5.6=16.8\)
21 tháng 4 2020
Bài 4:
a)
\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\\ \Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}-\frac{7x^2-14x-5}{15}=0\\ \Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\\ \Leftrightarrow36x+3=0\\ \Rightarrow x=-\frac{3}{36}==-\frac{1}{12}\)
b)
\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\\ \Leftrightarrow\frac{8x^2-32x+32}{24}-\frac{12x^2-27}{24}+\frac{4x^2-32x+64}{24}=0\\ \Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\\ \Leftrightarrow96-64x=0\\ \Rightarrow x=\frac{96}{64}=\frac{3}{2}\)
21 tháng 4 2020
Bài 3 câu g:
\(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{20}\)
\(\Leftrightarrow\left(\frac{x-10}{1994}-1\right)+\left(\frac{x-8}{1996}-1\right)+\left(\frac{x-6}{1998}-1\right)+\left(\frac{x-4}{2000}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2000}{4}-1\right)+\left(\frac{x-1998}{6}-1\right)+\left(\frac{x-1996}{8}-1\right)+\left(\frac{x-1994}{10}-1\right)\)
\(\Leftrightarrow\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}\right)=\left(x-2004\right)\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}\right)\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)\)
\(\Rightarrow x-2004=0\\ \Rightarrow x=2004\)
27 tháng 4 2020
e)
\(\left(x^2-1\right)\cdot\left(x^2+4x+3\right)=45\\ \Leftrightarrow x^4+4x^3+2x^2-4x-48=0\\ \Leftrightarrow\left(x^3+6x^2+14x+24\right)\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+2x+6\right)\cdot\left(x+4\right)\cdot\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
27 tháng 4 2020
d)
\(\left(x^2+3x+2\right)\cdot\left(x^2+5x+6\right)=72\\ \Leftrightarrow x^4+8x^3+23x^2+28x-60=0\\ \Leftrightarrow\left(x^3+9x^2+32x+60\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+4x+12\right)\cdot\left(x+5\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
15 tháng 2 2017
Bài này dà quá, violympic mà kết quả nè 101
P(-1)=2-3+4-5+....+198-199+200(Cái này dựa theo đề, nhân vô nó ra )
P(-1)=-1.99(99 cặp, t không tính) +200=101
28 tháng 4 2020
Bài 3:
a)
\(\left(x^2-5x\right)^2+10\cdot\left(x^2-5x\right)+24=0\\ \Leftrightarrow x\cdot\left(x-5\right)^2+10x\cdot\left(x-5\right)+24=0\\ \Leftrightarrow x^4-10x^3+35x^2-50x+24=0\\ \Leftrightarrow x^4-x^3-9x^3+9x^2+26x^2-26x-24x+24=0\\ \Leftrightarrow x^3\cdot\left(x-1\right)-9x^2\cdot\left(x-1\right)+26x\cdot\left(x-1\right)-24\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x^3-9x^2+26x-24\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x^3-2x^2-7x^2+14x+12x-24\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left[x^2\cdot\left(x-2\right)-7x\cdot\left(x-2\right)+12\cdot\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x^2-7x+12\right)=0\\ \)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x^2-3x-4x+12\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left[x\cdot\left(x-3\right)-4\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
b)
\(x\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)=42\\ \Leftrightarrow x^4+2x^3+2x^2+x-42=0\\ \Leftrightarrow x^4-2x^3+4x^3-8x^2+10x^2-20x+21x-42=0\\ \Leftrightarrow x^3\cdot\left(x-2\right)+4x^2\cdot\left(x-2\right)+10x\cdot\left(x-2\right)+21\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x^3+4x^2+10x+21\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x^3+3x^2+x^2+3x+7x+21\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left[x^2\cdot\left(x+3\right)+x\cdot\left(x+3\right)+7\cdot\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+3\right)\cdot\left(x^2+x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c)
\(\left(5x^2+3x-2\right)^2-\left(4x^2-x-5\right)^2=0\\ \Leftrightarrow\left(5x^2+3x-2+4x^2-x-5\right)\cdot\left(5x^2+3x-2-4x^2+x+5\right)=0\\ \Leftrightarrow\left(9x^2+2x-7\right)\cdot\left(x^2+4x+3\right)=0\\ \Leftrightarrow\left(9x^2+9x-7x-7\right)\cdot\left(x^2+3x+x+3\right)=0\\ \Leftrightarrow\left[9x\cdot\left(x+1\right)-7\cdot\left(x+1\right)\right]\cdot\left[x\cdot\left(x+3\right)+\left(x+3\right)\right]=0\\ \Leftrightarrow\left(9x-7\right)\cdot\left(x+1\right)\cdot\left(x+3\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\\9x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-3\\x=\frac{7}{9}\end{matrix}\right.\)
27 tháng 4 2020
Bài 2:
a)
\(x^2-6x+9=49\\ \Leftrightarrow x^2-6x+9-49=0\\ \Leftrightarrow x^2-6x-40=0\\ \Leftrightarrow x^2+4x-10x-40=0\\ \Leftrightarrow x\cdot\left(x+4\right)-10\cdot\left(x+4\right)=0\\ \Leftrightarrow\left(x-10\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-10=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
b)
\(x^3-2x^2-x+2=0\\ \Leftrightarrow x^2\cdot\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-1\right)\cdot\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot\left(x+1\right)\cdot\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
c)
\(x^3-3x^2-6x+8=0\\ \Leftrightarrow x^3-2x^2-8x-x^2+2x+8=0\\ \Leftrightarrow x^2\cdot\left(x-1\right)-2x\cdot\left(x-1\right)-8\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-2x-8\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+2x-4x-8\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left[x\cdot\left(x+2\right)-4\cdot\left(x+2\right)\right]\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-4\right)\cdot\left(x+2\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-2\\x=1\end{matrix}\right.\)
d)
\(x^4+2x^3+5x^2+4x-12=0\\ \Leftrightarrow x^4+3x^3+8x^2+12x-x^3-3x^2-8x-12=0\\ \Leftrightarrow x^3\cdot\left(x-1\right)+3x^2\cdot\left(x-1\right)+8x\cdot\left(x-1\right)+12\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+x^2+6x+2x^2+2x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left[x^2\cdot\left(x+2\right)+x\cdot\left(x+2\right)+6\cdot\left(x+2\right)\right]\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\cdot\left(x+2\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
TT
2
30 tháng 4 2017
đề 1 bài 4
xét tam gics ABC và tam giác HBA có
góc B chung
góc BAC = góc BHA (=90 độ)
=> tam giác ABC đồng dạng vs tam giác HBA (g.g)
=> AB/HB=BC/AB=> AB^2=HB *BC
áp dụng đl py ta go trog tam giác vuông ABC có
BC^2 = AB^2 +AC^2=6^2+8^2=100
=> BC =\(\sqrt{100}\)=10 cm
ta có tam giác ABC đồng dạng vs tam giác HBA (cm câu a )
=> AC/AH=BC/BA=>AH=8*6/10=4.8CM
=>AB/BH=AC/AH=> BH=6*4.8/8=3,6cm
=>HC =BC-BH=10-3,6=6,4cm
30 tháng 4 2017
dề 1 bài 1
5x+12=3x -14
<=>5x-3x=-14-12
<=>2x=-26
<=> x=-12
vạy S={-12}
(4x-2)*(3x+4)=0
<=>4x-2=0<=>x=1/2
<=>3x+4=0<=>x=-4/3
vậy S={1/2;-4/3}
đkxđ : x\(\ne2;x\ne-3\)
\(\dfrac{4}{x-2}+\dfrac{1}{x+3}=0\)
<=> 4(x+3)/(x-2)(x+3)+1(x-2)/(x-2)(x+3)
=> 4x+12+x-2=0
<=>5x=-10
<=>x=-2 (nhận)
vậy S={-2}
3 tháng 3 2017
\(x+y+z=0\Leftrightarrow x^2+y^2+z^2=-2\left(xy+xz+yz\right)\)
Mẫu số nhân ra : \(2\left(x^2+y^2+z^2\right)-2\left(xy+xz+yz\right)=3\left(x^2+y^2+z^2\right)\)
\(A=\dfrac{18\left(x^2+y^2+z^2\right)}{3\left(x^2+y^2+z^2\right)}=6\)
Bài 1:
6, x - \(\frac{x+1}{3}\) = \(\frac{2x+1}{5}\)
\(\Leftrightarrow\) \(\frac{15x}{15}\) - \(\frac{5\left(x+1\right)}{15}\) = \(\frac{3\left(2x+1\right)}{15}\)
\(\Leftrightarrow\) 15x - 5(x + 1) = 3(2x + 1)
\(\Leftrightarrow\) 15x - 5x - 5 = 6x + 3
\(\Leftrightarrow\) 10x - 5 = 6x + 3
\(\Leftrightarrow\) 10x - 6x = 3 + 5
\(\Leftrightarrow\) 4x = 8
\(\Leftrightarrow\) x = 2
Vậy S = {2}
làm lỗi nên hơi lâu
Chúc bạn học tốt!
1) \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\)
\(\Leftrightarrow\frac{9x+6}{6}-\frac{3x+1}{6}-\frac{10}{6}-\frac{12x}{6}=0\)
\(\Leftrightarrow\frac{9x+6-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow\frac{-6x-5}{6}=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\Leftrightarrow x=-\frac{5}{6}\)
Vậy \(S=\left\{-\frac{5}{6}\right\}\)
2) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
\(\Leftrightarrow\frac{3x-9}{15}-\frac{90}{15}+\frac{5-10x}{15}=0\)
\(\Leftrightarrow3x-9-90+5-10x=0\)
\(\Leftrightarrow-7x-94=0\)
\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)
Vậy \(S=-\frac{94}{7}\)
Chỗ câu 2, kết luận bị thiếu nha! Viết lại:
Vậy \(S=\left\{-\frac{94}{7}\right\}\)
7, \(\frac{3\left(x-3\right)}{4}\) + \(\frac{4x-10,5}{10}\) = \(\frac{3\left(x+1\right)}{5}\) + 6
\(\Leftrightarrow\) \(\frac{15\left(x-3\right)}{20}\) + \(\frac{2\left(4x-10,5\right)}{20}\) = \(\frac{12\left(x+1\right)}{20}\) + \(\frac{120}{20}\)
\(\Leftrightarrow\) 15(x - 3) + 2(4x - 10,5) = 12(x + 1) + 120
\(\Leftrightarrow\) 15x - 45 + 8x - 21 = 12x + 12 + 120
\(\Leftrightarrow\) 23x - 66 = 12x + 132
\(\Leftrightarrow\) 23x - 12x = 132 + 66
\(\Leftrightarrow\) 11x = 198
\(\Leftrightarrow\) x = 18
Vậy S = {18}
Chúc bạn học tốt!
Sợ bạn ko đọc được nên mk làm từng câu cho nhanh :))
3) \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
\(\Leftrightarrow2x+\frac{6}{5}-5+\frac{13}{5}-x=0\)
\(\Leftrightarrow\frac{10x}{5}+\frac{6}{5}-\frac{25}{5}+\frac{13}{5}-\frac{5x}{5}=0\)
\(\Leftrightarrow10x+6-25+13-5x=0\)
\(\Leftrightarrow5x-6=0\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)
Vậy \(S=\left\{\frac{6}{5}\right\}\)
Nam: đăng lộn chỗ, lỡ đăng vào phần bl, xin lỗi nhé!
Đó là câu 3
8, \(\frac{2\left(3x+1\right)+1}{4}\) - 5 = \(\frac{2\left(3x-1\right)}{5}\) - \(\frac{3x+2}{10}\)
\(\Leftrightarrow\) \(\frac{5\left[2\left(3x+1\right)+1\right]}{20}-\frac{100}{20}=\frac{8\left(3x-1\right)}{20}-\frac{2\left(3x+2\right)}{20}\)
\(\Leftrightarrow\) 5[2(3x + 1) + 1] - 100 = 8(3x - 1) - 2(3x + 2)
\(\Leftrightarrow\) 5(6x + 2 + 1) - 100 = 24x - 8 - 6x - 4
\(\Leftrightarrow\) 30x + 10 + 5 - 100 = 24x - 8 - 6x - 4
\(\Leftrightarrow\) 30x - 85 = 18x - 12
\(\Leftrightarrow\) 30x - 18x = -12 + 85
\(\Leftrightarrow\) 12x = 73
\(\Leftrightarrow\) x = \(\frac{73}{12}\)
Vậy S = {\(\frac{73}{12}\)}
4)\(\frac{2x+3}{3}=\frac{5-4x}{2}\)
\(\Leftrightarrow\frac{4x+6}{6}=\frac{15-12x}{6}\)
\(\Leftrightarrow4x+6-15+12x=0\)
\(\Leftrightarrow16x-9=0\)
\(\Leftrightarrow16x=9\Leftrightarrow x=\frac{9}{16}\)
Vậy \(S=\left\{\frac{9}{16}\right\}\)
5) \(\frac{5x+3}{12}=\frac{1+2x}{9}\)
<=>\(\frac{15x+9}{36}=\frac{4+8x}{36}\)
<=> \(15x+9-4-8x=0\)
<=> \(7x+5=0\Leftrightarrow x=-\frac{5}{7}\)
Vậy \(S=\left\{-\frac{5}{7}\right\}\)
9, \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
\(\Leftrightarrow\) \(\frac{4\left(x+1\right)}{12}+\frac{9\left(2x+1\right)}{12}=\frac{2\left[2x+3\left(x+1\right)\right]}{12}+\frac{7+12x}{12}\)
\(\Leftrightarrow\) 4(x + 1) + 9(2x + 1) = 2[2x + 3(x + 1)] + 7 + 12x
\(\Leftrightarrow\) 4x + 4 + 18x + 9 = 2(2x + 3x + 3) + 7 + 12x
\(\Leftrightarrow\) 22x + 13 = 4x + 6x + 6 + 7 + 12x
\(\Leftrightarrow\) 22x + 13 = 22x + 13
\(\Leftrightarrow\) 22x - 22x = 13 - 13
\(\Leftrightarrow\) 0x = 0
\(\Rightarrow\) Phương trình đúng với mọi x
Vậy phương trình đúng với mọi x
10, \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\) \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21x}{21}+\frac{273}{21}\)
\(\Leftrightarrow\) 7(2x - 1) - 3(5x + 2) = 21x + 273
\(\Leftrightarrow\) 14x - 7 - 15x - 6 = 21x + 273
\(\Leftrightarrow\) -x - 13 = 21x + 273
\(\Leftrightarrow\) -x - 21x = 273 + 13
\(\Leftrightarrow\) -22x = 286
\(\Leftrightarrow\) x = -13
Vậy S = {-13}
Chúc bạn học tốt!
câu 3 của Trang với câu 6 của tui là bài 1
câu 1 với câu 2 là bài 1
Câu 7 là bài 1
Câu 10 là bài 1
Bài 2:
11) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\)
\(\Leftrightarrow20x-20=0\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
Bài 2:
1, \(\frac{7x-7}{x-1}=\frac{2}{3}\)
\(\Leftrightarrow\) \(\frac{3\left(7x-7\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Leftrightarrow\) 3(7x - 7) = 2(x - 1)
\(\Leftrightarrow\) 21x - 21 = 2x - 2
\(\Leftrightarrow\) 21x - 2x = -2 + 21
\(\Leftrightarrow\) 19x = 19
\(\Leftrightarrow\) x = 1
Vậy S = {1}
2, \(\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow\) \(\frac{4\left(3-7x\right)}{2\left(1+x\right)}=\frac{1+x}{2\left(1+x\right)}\)
\(\Leftrightarrow\) 4(3 - 7x) = 1 + x
\(\Leftrightarrow\) 12 - 28x = 1 + x
\(\Leftrightarrow\) -28x - x = 1 - 12
\(\Leftrightarrow\) -29x = -11
\(\Leftrightarrow\) x = \(\frac{11}{29}\)
Vậy S = {\(\frac{11}{29}\)}
Câu 2:
3, \(\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
\(\Leftrightarrow\) \(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)
\(\Leftrightarrow\) 1 + 3(x - 2) = 3 - x
\(\Leftrightarrow\) 1 + 3x - 6 = 3 - x
\(\Leftrightarrow\) 3x - 5 = 3 - x
\(\Leftrightarrow\) 3x + x = 3 + 5
\(\Leftrightarrow\) 4x = 8
\(\Leftrightarrow\) x = 2
Vậy S = {2}
Câu 2
4, \(\frac{2}{x+1}-\frac{1}{x-1}=\frac{3}{x^2-1}\)
\(\Leftrightarrow\) \(\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=\frac{3}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\) 2(x - 1) - x - 1 = 3
\(\Leftrightarrow\) 2x - 2 - x - 1 = 3
\(\Leftrightarrow\) x - 3 = 3
\(\Leftrightarrow\) x = 3
Vậy S = {3}
Bài 2:
12) \(5+\frac{76}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{ }}}}}}}}}}}}}}}\)
\(\Leftrightarrow\frac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{76}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow5\left(x+4\right)\left(x-4\right)+76-\left(2x-1\right)\left(x-4\right)-\left(3x-1\right)\left(x+4\right)=0\)
\(\Leftrightarrow5x^2-80+76-2x^2+9x-4-3x^2-11x+4=0\)
\(\Leftrightarrow-2x-4=0\Leftrightarrow x=-2\)
Vậy \(S=\left\{-2\right\}\)
#Cũng không chắc lắm, nếu sai thì nói với mình!
Bài 2:
13) ĐKXĐ: \(x\ne\frac{2}{3};x\ne-\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\)
\(\Leftrightarrow16-6x=0\)
\(\Leftrightarrow6x=16\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)
Vậy \(S=\left\{\frac{8}{3}\right\}\)
14) ĐKXĐ: \(x\ne\frac{1}{4};x\ne-\frac{1}{4}\)
\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8x+6}{16x^2-1}\)
\(\Leftrightarrow\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8x+6}{\left(4x-1\right)\left(4x+1\right)}=0\)
\(\Leftrightarrow-12x-3-8x+2+8x+6=0\)
\(\Leftrightarrow5-12x=0\)
\(\Leftrightarrow12x=5\Leftrightarrow x=\frac{5}{12}\left(TM\right)\)
Vậy \(S=\left\{\frac{5}{12}\right\}\)
Bài 2:
15, \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\) \(\frac{3\left(3-5x\right)\left(x-3\right)}{\left(5x-1\right)\left(3-5x\right)\left(x-3\right)}+\frac{2\left(5x-1\right)\left(x-3\right)}{\left(5x-1\right)\left(3-5x\right)\left(x-3\right)}=\frac{-4\left(3-5x\right)}{\left(5x-1\right)\left(3-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\) 3(3 - 5x)(x - 3) + 2(5x - 1)(x - 3) = -4(3 - 5x)
\(\Leftrightarrow\) −15x2 + 54x − 27 + 10x2 − 32x + 6 = -12 + 20x
\(\Leftrightarrow\) −5x2 + 22x − 21 = -12 + 20x
\(\Leftrightarrow\) −5x2 + 22x - 20x = -12 + 21
\(\Leftrightarrow\) −5x2 + 2x = 9
\(\Leftrightarrow\) −5x2 + 2x − 9 = 0
\(\Leftrightarrow\) −5x2 + 5x - 3x - 9 = 0
\(\Leftrightarrow\) −5x(x + 1) - 3(x + 3) = 0
Hình như đề lỗi rồi, phân tích như ở trên cx ko ra được đâu vì kết quả là:
x = \(\frac{1}{5}\) hoặc x = \(\frac{1+2x\sqrt{11}}{5}\) hoặc x = \(\frac{1-2x\sqrt{11}}{5}\)
Bạn xem lại đề đi! (câu 15 ấy)
Bài 2:
5, \(\frac{1}{x-1}+\frac{2}{x+1}=\frac{x}{x^2-1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\) x + 1 + 2(x - 1) = x
\(\Leftrightarrow\) x + 1 + 2x - 2 = x
\(\Leftrightarrow\) 3x - 1 = x
\(\Leftrightarrow\) 3x - x = 1
\(\Leftrightarrow\) 2x = 1
\(\Leftrightarrow\) x = \(\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
6, \(\frac{x}{x-1}-\frac{2}{x^2-1}=0\) (ĐKXĐ: x - 1 \(\ne\) 0 và x2 - 1 \(\ne\) 0 \(\Rightarrow\) x \(\ne\) 1 và x \(\ne\) -1)
\(\Leftrightarrow\) \(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\) x(x + 1) - 2 = 0
\(\Leftrightarrow\) x2 + x - 2 = 0
\(\Leftrightarrow\) x2 + x + \(\frac{1}{4}\) - \(\frac{9}{4}\) = 0
\(\Leftrightarrow\) (x + \(\frac{1}{2}\))2 - \(\frac{9}{4}\) = 0
\(\Leftrightarrow\) x + \(\frac{1}{2}\) = \(\frac{3}{2}\) hoặc x + \(\frac{1}{2}\) = -\(\frac{3}{2}\)
\(\Leftrightarrow\) x = 1 hoặc x = -2
Vậy S = {1; -2}
7, \(\frac{x}{x-2}-\frac{2x}{x+2}=\frac{5}{x^2-4}\)
\(\Leftrightarrow\) \(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{5}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\) x(x + 2) - 2x(x - 2) = 5
\(\Leftrightarrow\) x2 + 2x - 2x2 + 4x = 5
\(\Leftrightarrow\) -x2 + 6x = 5
\(\Leftrightarrow\) -x2 + 6x - 5 = 0
\(\Leftrightarrow\) -x2 + x + 5x - 5 = 0
\(\Leftrightarrow\) -x(x - 1) + 5(x - 1) = 0
\(\Leftrightarrow\) (x - 1)(-x + 5) = 0
\(\Leftrightarrow\) x - 1 = 0 hoặc -x + 5 = 0
\(\Leftrightarrow\) x = 1 hoặc x = 5
Vậy S = {1; 5}
Bài 2:
17, \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\) \(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\) x2 + x + 1 - 3x2 = 2x(x - 1)
\(\Leftrightarrow\) -2x2 + x + 1 = 2x2 - 2x
\(\Leftrightarrow\) -2x2 - 2x2 + x + 2x = -1
\(\Leftrightarrow\) -4x2 + 3x = -1
\(\Leftrightarrow\) x(-4x + 3) = -1
\(\Leftrightarrow\) x = 1
Vậy S = {1}
19, \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\) (ĐKXĐ là mẫu nha)
\(\Leftrightarrow\) \(\frac{x+1}{2x^2-7x+3}-\frac{2x+5}{2x^2-7x+3}+\frac{x+4}{2x^2-5x+2}=0\)
\(\Leftrightarrow\) \(\frac{x+1-2x-5}{2x^2-7x+3}+\frac{x+4}{2x^2-5x+2}=0\)
\(\Leftrightarrow\) \(\frac{-4-x}{2x^2-7x+3}+\frac{x+4}{2x^2-5x+2}=0\)
\(\Leftrightarrow\) \(\frac{-\left(4+x\right)}{2x^2-7x+3}+\frac{x+4}{2x^2-5x+2}=0\)
\(\Leftrightarrow\) -(4 + x)(\(\frac{1}{2x^2-7x+3}\)) + (x + 4)(\(\frac{1}{2x^2-5x+2}\)) = 0
\(\Leftrightarrow\) (x + 4)(\(\frac{-1}{2x^2-7x+3}\) + \(\frac{1}{2x^2-5x+2}\)) = 0
\(\Leftrightarrow\) x + 4 = 0
\(\Leftrightarrow\) x = -4
Vậy S = {-4}
Bài 2:
18) ĐKXĐ: \(x\ne1\)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=1\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)