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\(2x^2+7x-9=0\Leftrightarrow\left(x-1\right)\left(x+\dfrac{9}{2}\right)=0\)
\(x\in\left\{1;\dfrac{-9}{2}\right\}\)
<=> 2x2 -2x +9x -9 =0
<=> 2x(x-1) + 9 (x-1) = 0
<=> (2x+9)(x-1) = 0
<=> 2x + 9 =0 hoặc x-1 = 0
<=> x= -9/2 hoặc x= 1
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\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
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\(a,6\left(x-2\right)=8\left(3x+1\right)\\ \Leftrightarrow6x-12=24x+8\\ \Leftrightarrow18x+20=0\\ \Leftrightarrow x=-\dfrac{10}{9}\\ b,2x-\left(3-7x\right)=5\left(x+3\right)\\ \Leftrightarrow2x-3+7x=5x+15\\ \Leftrightarrow9x-3-5x-15=0\\ \Leftrightarrow4x-18=0\\ \Leftrightarrow x=\dfrac{9}{2}\\ c,\left(x-1\right)^2=\left(x+3\right)\left(x+2\right)\\ \Leftrightarrow x^2-2x+1=x^2+5x+6\\ \Leftrightarrow7x+5=0\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d,\left(3x-9\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(e,x^2-3x+2=0\\ \Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\\ \Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ f,x^2-4x+4=0\\ \Leftrightarrow x^2-2.2+2^2=0\\ \Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ x=2\)
a, \(6x-12=24x+8\Leftrightarrow18x=-20\Leftrightarrow x=-\dfrac{20}{18}=-\dfrac{10}{9}\)
b, \(2x-3+7x=5x+15\Leftrightarrow4x=18\Leftrightarrow x=\dfrac{9}{2}\)
c, \(x^2-2x+1=x^2+5x+6\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)
d, \(\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)
e, \(x^2-3x+2=0\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow x=1;x=2\)
f, \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
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\(=\dfrac{5x}{2\left(x-2\right)}\cdot\dfrac{x-2}{3x}=\dfrac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)
\(x\left(x-2\right)-\left(7x-14\right)=0\\ \Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
x ( x - 2 ) - ( 7x - 14 ) = 0
x ( x - 2 ) - 7 ( x - 2 ) = 0
( x - 2 ) ( x - 7 ) = 0
TH1: x - 2 = 0 TH2: x - 7 = 0
=> x = 2 => x = 7
Vậy x = 2 hoặc x = 7
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