Bài 6: Tìm x, biết: a) x(x – 2) + x – 2 = 0 b) (2x – 1) ² – (x + 3) ² = 0 c) 2x(x – 3) + 5(3 – x) = 0 d) (x ² – 1) – (x – 1)(3 – 2x) = 0 e) 9(x – 2) ²

$a)_{}$ $x.(x-2)+x-2=0_{}$

$⇔x.(x-2)+(x-2)=0_{}$

$⇔(x+1)(x-2)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\) 

$b)_{}$ $(2x-1)^2-(x+3)^2=0_{}$

$⇔4x^2-4x+1-(x^2+6x+9)=0_{}$

$⇔4x^2-4x+1-x^2-6x-9=0_{}$

$⇔3x^2-10x+8=0_{}$

$⇔3x^2+2x-12x-8=0_{}$

$⇔x.(3x+2)-4.(3x+2)=0_{}$

$⇔(3x+2).(x-4)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}3x+2=0\\x-4=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-\frac{2}{3}\\x=4\end{array} \right.\) 

$c)_{}$ $2x.(x-3)+5.(3-x)=0_{}$

$⇔2x.(x-3)-5.(x-3)=0_{}$

$⇔(2x-5)(x-3)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}2x-5=0\\x-3=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=3\end{array} \right.\)

$d)_{}$ $(x^2-1)-(x-1)(3-2x)=0_{}$

$⇔(x-1)^2+(-x+1).(3-2x)=0_{}$

$⇔(x-1)^2-(x-1)(3-2x)_{}$

$⇔(x-1)^2. [ x+1-(3-2x)]=0_{}$

$⇔(x-1)(x+1-3+2x)=0_{}$

$⇔(x-1)(3x-2)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}x-1=0\\3x-4=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=1\\x=\frac{2}{3}\end{array} \right.\)

$e)_{}$ $9.(x-2)^2=4.(1-2x)^2_{}$

$⇔9.(x^2-4x+4)=4.(1-4x+4x^2)_{}$

$⇔9x^2-36x+36=4-16x+16x^2_{}$

$⇔9x^2-36x+36-4+16x-16x^2=0_{}$

$⇔-7x^2-20x+32=0_{}$

$⇔7x^2+20x-32=0_{}$

$⇔7x^2+28x-8x-32=0_{}$

$⇔7x.(x+4)-8.(x+4)=0_{}$

$⇔(x+4).(7x-8)=0_{}$

$⇔_{}$ \(\left[ \begin{array}{l}x+4=0\\7x-8=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-4\\x=\frac{8}{7}\end{array} \right.\)

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