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

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)

\(\left(5x^2+3x-2\right)^2-\left(4x^2-x-5\right)^2=0\\ \Leftrightarrow\left(5x^2+3x-2+4x^2-x-5\right)\cdot\left(5x^2+3x-2-4x^2+x+5\right)=0\\ \Leftrightarrow\left(9x^2+2x-7\right)\cdot\left(x^2+4x+3\right)=0\\ \Leftrightarrow\left(9x^2+9x-7x-7\right)\cdot\left(x^2+3x+x+3\right)=0\\ \Leftrightarrow\left[9x\cdot\left(x+1\right)-7\cdot\left(x+1\right)\right]\cdot\left[x\cdot\left(x+3\right)+\left(x+3\right)\right]=0\\ \Leftrightarrow\left(9x-7\right)\cdot\left(x+1\right)\cdot\left(x+3\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\\9x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=-3\\x=\frac{7}{9}\end{matrix}\right.\)

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)

\(x^4+2x^3+5x^2+4x-12=0\\ \Leftrightarrow x^4+3x^3+8x^2+12x-x^3-3x^2-8x-12=0\\ \Leftrightarrow x^3\cdot\left(x-1\right)+3x^2\cdot\left(x-1\right)+8x\cdot\left(x-1\right)+12\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^3+x^2+6x+2x^2+2x+12\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left[x^2\cdot\left(x+2\right)+x\cdot\left(x+2\right)+6\cdot\left(x+2\right)\right]\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\cdot\left(x+2\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

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

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}

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 \(S=-\frac{94}{7}\)

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 \(NC=8\\ BC=15\\ MN=5\\ BN=9.8\)

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:

\(AB+AC+BC=4.2+7+5.6=16.8\)

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)

\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\cdot\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\\ \Leftrightarrow\frac{8x^2-32x+32}{24}-\frac{12x^2-27}{24}+\frac{4x^2-32x+64}{24}=0\\ \Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\\ \Leftrightarrow96-64x=0\\ \Rightarrow x=\frac{96}{64}=\frac{3}{2}\)

Bài 3 câu g:

\(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{20}\)

\(\Leftrightarrow\left(\frac{x-10}{1994}-1\right)+\left(\frac{x-8}{1996}-1\right)+\left(\frac{x-6}{1998}-1\right)+\left(\frac{x-4}{2000}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2000}{4}-1\right)+\left(\frac{x-1998}{6}-1\right)+\left(\frac{x-1996}{8}-1\right)+\left(\frac{x-1994}{10}-1\right)\)

\(\Leftrightarrow\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}\right)=\left(x-2004\right)\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}\right)\)

\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)\)

\(\Rightarrow x-2004=0\\ \Rightarrow x=2004\)

Nam ơi phần b trên cùng có phải giúp ko :))

Làm đi cho bớt rảnh :))

\(x+y+z=0\Leftrightarrow x^2+y^2+z^2=-2\left(xy+xz+yz\right)\)

Mẫu số nhân ra : \(2\left(x^2+y^2+z^2\right)-2\left(xy+xz+yz\right)=3\left(x^2+y^2+z^2\right)\)

\(A=\dfrac{18\left(x^2+y^2+z^2\right)}{3\left(x^2+y^2+z^2\right)}=6\)