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\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)
\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
PS: Điều kiện xác đinh bạn tự làm nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
d: Ta có: \(9x^2+6x-8=0\)
\(\Leftrightarrow9x^2+12x-6x-8=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
e: Ta có: \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f: Ta có: \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
hay \(x\in\left\{0;-3\right\}\)
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a ) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
b ) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)=0\)
\(\Leftrightarrow x=3\)
a) x3 - 6x2 + 12x - 8 = 0
( x - 2 ) 3 = 0
x - 2 = 0
x = 2
b) x3 + 9x2 + 27x + 27 = 0
( x + 3 )3 = 0
x + 3 = 0
x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(9x^2-6x-8=0\)
\(x.\left(9x-6\right)=8\)
Lập bảng xét là xong ok
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9x^2+6x-8=0\)
\(\Leftrightarrow9x^2-6x+12x-8=0\)
\(\Leftrightarrow3x\left(3x-2\right)+4\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...
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1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
9x2 +6x-8=0
<=> 9x2 +6x+1-9=0
<=> (3x+1)^2 - 3^2 = 0
<=> (3x+1-3)(3x+1+3)=0
<=> (3x-2)(3x+4)=0
=> TH1: 3x-2=0 <=> x=2/3
TH2: 3x+4=0 <=> x= -4/3
vậy ....................
\(9x^2+6x-8=0\)
\(9x^2+6x+1-9=0\)
\(\left(3x+1\right)^2-3^2=0\)
\(\left(3x+1-3\right)\left(3x+1+3\right)=0\)
\(\left(3x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\3x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-4}{3}\end{cases}}\)
vay \(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-4}{3}\end{cases}}\)