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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:
\(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{20}\)
\(\Leftrightarrow\left(\frac{x-10}{1994}-1\right)+\left(\frac{x-8}{1996}-1\right)+\left(\frac{x-6}{1998}-1\right)+\left(\frac{x-4}{2000}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2000}{4}-1\right)+\left(\frac{x-1998}{6}-1\right)+\left(\frac{x-1996}{8}-1\right)+\left(\frac{x-1994}{10}-1\right)\)
\(\Leftrightarrow\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}\right)=\left(x-2004\right)\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{10}\right)\)
\(\Leftrightarrow\left(x-2004\right)\cdot\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)\)
\(\Rightarrow x-2004=0\\ \Rightarrow x=2004\)
29 tháng 3 2020
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!
LT
29 tháng 3 2020
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 \(S=-\frac{94}{7}\)
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:
\(AB+AC+BC=4.2+7+5.6=16.8\)
LT
16 tháng 3 2020
Làm hết toàn bộ luôn à??? Thấy có vài câu khá đơn giản chắc làm đc đó! :))
TT
25 tháng 10 2017
Đề số 3.
1.
a,\(4x\left(5x^2-2x+3\right)\)
\(=20x^3-8x^2+12x\)
b.\(\left(x-2\right)\left(x^2-3x+5\right)\)
\(=x^3-3x^2+5x-2x^2+6x-10\)
\(=x^3-5x^2+11x-10\)
c,\(\left(10x^4-5x^3+3x^2\right):5x^2\)
\(=2x^2-x+\dfrac{3}{5}\)
d,\(\left(x^2-12xy+36y^2\right):\left(x-6y\right)\)
\(=\left(x-6y\right)^2:\left(x-6y\right)\)
\(=x-6y\)
2.
a,\(x^2+5x+5xy+25y\)
\(=\left(x^2+5x\right)+\left(5xy+25y\right)\)
\(=x\left(x+5\right)+5y\left(x+5\right)\)
\(=\left(x+5y\right)\left(x+5\right)\)
b,\(x^2-y^2+14x+49\)
\(=\left(x^2+14x+49\right)-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+7-y\right)\left(x+7+y\right)\)
c,\(x^2-24x-25\)
\(=x^2+25x-x-25\)
\(=\left(x^2-x\right)+\left(25x-25\right)\)
\(=x\left(x-1\right)+25\left(x-1\right)\)
\(=\left(x+25\right)\left(x-1\right)\)
3.
a,\(5x\left(x-3\right)-x+3=0\)
\(5x\left(x-3\right)-\left(x-3\right)=0\)
\(\left(5x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{5}\) hoặc \(x=3\)
b.\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(3x^2-15x-\left(2x+3x^2-2-3x\right)=30\)
\(3x^2-15x-2x-3x^2+2+3x=30\)
\(-14x+2=30\)
\(-14x=28\)
\(x=-2\)
c,\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\)
\(x^2+5x+6-x^2-5x+2x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
Mình học chật hình không giúp bạn được.Xin lỗi!
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)
\(\left(x^2+3x+2\right)\cdot\left(x^2+5x+6\right)=72\\ \Leftrightarrow x^4+8x^3+23x^2+28x-60=0\\ \Leftrightarrow\left(x^3+9x^2+32x+60\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2+4x+12\right)\cdot\left(x+5\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Bài 3:
1)
\(2x^2+5x+3=0\\ \Leftrightarrow\left(3+2x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+2x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=-1\end{matrix}\right.\)
2)
\(x^2+4x+3=0\\ \Leftrightarrow\left(3+x\right)\cdot\left(1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3+x=0\\1+x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
3)
\(x^2-x-12=0\\ \Leftrightarrow\left(-3-x\right)\cdot\left(4-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-x=0\\4-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
4)
\(x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
5)
\(-x^2+5x-6=0\\ \Leftrightarrow\left(-3+x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3+x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
6)
\(4x^2-12x+5=0\\ \Leftrightarrow\left(1-2x\right)\cdot\left(5-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-2x=0\\5-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
7)
\(4x^2+4x-3=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
8)
\(x^2-3x+2=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
9)
\(3x^2-22x-16=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.\)
10)
\(2x^2+7x-15=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
11)
\(\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
12)
\(\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
13)
\(25-\left(3-x\right)^2=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.\)
14)
\(\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1\)
Bài 3
1, 2x2 + 5x + 3 = 0
\(\Leftrightarrow\) 2x2 + 2x + 3x + 3 = 0
\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(2x + 3) = 0
\(\Leftrightarrow\) x + 1 = 0 hoặc 2x + 3 = 0
\(\Leftrightarrow\) x = -1 và x = \(\frac{-3}{2}\)
Vậy S = {-1; \(\frac{-3}{2}\)}
2, x2 + 4x + 3 = 0
\(\Leftrightarrow\) x2 + x + 3x + 3 = 0
\(\Leftrightarrow\) x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(x + 3) = 0
\(\Leftrightarrow\) x + 1 = 0 hoặc x + 3 = 0
\(\Leftrightarrow\) x = -1 hoặc x = -3
Vậy S = {-1; -3}
tiếp bài 3
14, (x - 3)2 - (x + 1)2 = 0
\(\Leftrightarrow\) (x - 3 - x - 1)(x - 3 + x + 1) = 0
\(\Leftrightarrow\) (-4)(2x - 2) = 0
\(\Leftrightarrow\) (-8)(x - 1) = 0
\(\Leftrightarrow\) x - 1 = 0
\(\Leftrightarrow\) x = 1
Vậy S = {1}
ok
Bài 3 (thêm bước):
3)
\(x^2-x-12=0\\ \Leftrightarrow-12+3x-4x+x^2=0\\ \Leftrightarrow-3\cdot\left(4-x\right)-x\cdot\left(4-x\right)=0\\ \Leftrightarrow\left(4-x\right)\cdot\left(-3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}4-x=0\\-3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
4)
\(x^2-3x+2=0\\ \Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(1-x\right)\cdot\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
5)
\(-x^2+5x-6=0\\ \Leftrightarrow-x^2+2x+3x-6=0\\ \Leftrightarrow x\cdot\left(2-x\right)-3\cdot\left(2-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
6)
\(4x^2-12x+5=0\\ \Leftrightarrow5-2x-10x+4x^2=0\\ \Leftrightarrow5\cdot\left(1-2x\right)-2x\cdot\left(1-2x\right)=0\\ \Leftrightarrow\left(5-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}5-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
7)
\(4x^2+4x-3=0\\ \Leftrightarrow-3+6x-2x+4x^2=0\\ \Leftrightarrow-3\cdot\left(1-2x\right)-2x\cdot\left(1-2x\right)=0\\ \Leftrightarrow\left(-3-2x\right)\cdot\left(1-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3-2x=0\\1-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
8)
\(x^2-3x+2=0\\ \Leftrightarrow2-3x+x^2=0\\ \Leftrightarrow2-x-2x+x^2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
9)
\(3x^2-22x-16=0\\ \Leftrightarrow-16-22x+3x^2=0\\ \Leftrightarrow-16+2x-24x+3x^2=0\\ \Leftrightarrow-2\cdot\left(8-x\right)-3x\cdot\left(8-x\right)=0\\ \Leftrightarrow\left(-2-3x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2-3x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=8\end{matrix}\right.\)
10)
\(2x^2+7x-15=0\\ \Leftrightarrow-15+7x+2x^2=0\\ \Leftrightarrow-15+10x-30+2x^2=0\\ \Leftrightarrow-2\cdot\left(3-2x\right)-x\cdot\left(3-2x\right)=0\\ \Leftrightarrow\left(-5-x\right)\cdot\left(3-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-5-x=0\\3-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
11)
\(\left(x-5\right)^2-16=0\\ \Leftrightarrow\left(x-5\right)^2-4^2=0\\ \Leftrightarrow\left(x-5-4\right)\cdot\left(x-5+4\right)=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
12)
\(\left(x-4\right)^2-25=0\\ \Leftrightarrow\left(x-4\right)^2-5^2=0\\ \Leftrightarrow\left(x-4-5\right)\cdot\left(x-4+5\right)=0\\ \Leftrightarrow\left(x-9\right)\cdot\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
13)
\(25-\left(3-x\right)^2=0\\ \Leftrightarrow5^2-\left(3-x\right)^2=0\\ \Leftrightarrow\left(5-3+x\right)\cdot\left(5+3-x\right)=0\\ \Leftrightarrow\left(2+x\right)\cdot\left(8-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2+x=0\\8-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=8\end{matrix}\right.\)
14)
\(\left(x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(x-3-x-1\right)\cdot\left(x-3+x+1\right)=0\\ \Leftrightarrow-4\cdot\left(2x-2\right)=0\\ \Rightarrow2x-2=0\\ \Rightarrow x=1\)
15)
\(\left(3x-7\right)^2-4\cdot\left(x+1\right)^2=0\\ \Leftrightarrow9x^2-42x+49-4x^2-8x-4=0\\ \Leftrightarrow5x^2-50x+45=0\\ \Leftrightarrow5\cdot\left(x^2-10x+9\right)=0\\ \Leftrightarrow5\cdot\left(9-9x-x+x^2\right)=0\\ \Leftrightarrow5\cdot\left[9\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\right]\\ \Leftrightarrow5\cdot\left(9-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}9-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
16)
\(\left(7x-4\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow\left(7x-4-2x-1\right)\cdot\left(7x-4+2x+1\right)=0\\ \Leftrightarrow\left(5x-5\right)\cdot\left(9x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-5=0\\9x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
17)
\(\left(x^2-4\right)\cdot\left(2x+3\right)=\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3\right)-\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3-x+1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(x+4\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
18)
\(x^3-25x=0\\ \Leftrightarrow x\cdot\left(x^2-25\right)=0\\ \Leftrightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
19)
\(x^3-16x=0\\ \Leftrightarrow x\cdot\left(x^2-16\right)=0\\ \Leftrightarrow x\cdot\left(x-4\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
20)
\(x^3-5x^2+6x=0\\ \Leftrightarrow x\cdot\left(x^2-5x+6\right)=0\\ \Leftrightarrow x\cdot\left(6-2x-3x+x^2=0\right)\\ \Leftrightarrow x\cdot\left[2\cdot\left(3-x\right)-x\cdot\left(3-x\right)=0\right]\\ \Leftrightarrow x\cdot\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Bài 5.
a)
Có AB < AC (vì n > m) (1)
Ta có: \(\frac{AB}{AC}=\frac{BD}{DC}\) ( vì AD là phân giác của góc BAC) (2)
Từ (1) và (2), ta có BD < CD
⇒ D nằm giữa B và M
Đặt S1, S2 lần lượt là diện tích △ADM và △ADC
Ta có:
\(\frac{S_1}{S_2}=\frac{\frac{1}{2}\cdot BD\cdot AH}{\frac{1}{2}\cdot CD\cdot AH}=\frac{BD}{CD}=\frac{AB}{AC}=\frac{m}{n}\)
⇒ \(\frac{S_1+S_2}{S_2}=\frac{m+n}{n}=\frac{S}{S_2}=\frac{m+n}{n}\Rightarrow S_2=\frac{n\cdot S}{m+n}\)
Vì \(S_{AMC}=S_{AMB}=\frac{1}{2}\cdot S\) ⇒ \(S_{ADM}\) là:
\(S_{ADM}=S_{ADC}-S_{AMC}=S_2-\frac{1}{2}\cdot S=\frac{n\cdot S}{m+n}-\frac{1}{2}\cdot S=\left[\frac{n-m}{2\cdot\left(m+n\right)}\right]\cdot S\)
b)
\(S_{ADM}=\left[\frac{7-3}{2\cdot\left(7+3\right)}\right]\cdot S=\frac{2}{10}\cdot S=\frac{1}{5}\cdot S=0.2\cdot S=20\%\cdot S\)
Vậy \(S_{ADM}=20\%\cdot S_{ABC}\)
Bài 4.
Bài 4.
a)Ta có \(\Delta ABC\) cân tại A nên: \(AB=AC=15\left(cm\right)\)
Mà BD là đường p/g đồng thời là đường trung tuyến ( AD = DC) nên:
\(AD=\frac{AC}{2}=\frac{15}{2}=7.5\left(cm\right)\\ \Rightarrow DC=AC=7.5\left(cm\right)\)
b)?
15)
\(\left(3x-7\right)^2-4\cdot\left(x+1\right)^2=0\\ \Leftrightarrow5x^2-50x+45\\ \Leftrightarrow5\cdot\left(1-x\right)\cdot\left(9-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}1-x=0\\9-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
16)
\(\left(7x-4\right)^2\cdot\left(2x+1\right)^2=0\\ \Leftrightarrow\left(5x-5\right)\cdot\left(9x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-5=0\\9x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
17)
\(\left(x^2-4\right)\cdot\left(2x+3\right)=\left(x^2-4\right)\cdot\left(x-1\right)\\ \Leftrightarrow\left(x^2-4\right)\cdot\left(2x+3\right)-\left(x^2-4\right)\cdot\left(x-1\right)=0\\ \Leftrightarrow\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
18)
\(x^3-25x=0\\ \Leftrightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
19)
\(x^3-16x=0\\ \Leftrightarrow x\cdot\left(x-4\right)\cdot\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
20)
\(x^3-5x^2+6x=0\\ \Leftrightarrow x\cdot\left(2-x\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2-x=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
bạn ơi hay bạn lm đề này đi đề này có bạn đang làm
https://hoc24.vn/hoi-dap/question/948593.html?auto=1
cảm ơn bạn
nhầm , đây bài 5, bài 4 mình đang bí