Đáp án `+` Giải thích các bước giải `!`
`to` Tìm `x:`
`a)`
`(x-2)^2-(x-2)(3x+2) = 0`
`<=> (x-2)(x-2-3x-2) = 0`
`<=> (x-2)(-2x-4) = 0`
`<=> -2(x-2)(x+2) = 0`
`<=> (x-2)(x+2) = 0`
`<=> x^2-4 = 0`
`<=> x^2 = 4`
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `S= {2; -2}`
`b)`
`(x-5)^2=9`
`<=> (x-5)^2 = (+-3)^2`
`<=>` \(\left[ \begin{array}{l}x-5=3\\x-5=-3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=8\\x=2\end{array} \right.\)
Vậy `S= {8; 2}`
`c)`
`2x^2+7x+6 = 0`
`<=> 2x^2+4x+3x+6 = 0`
`<=> (2x^2+4x)+(3x+6) = 0`
`<=> 2x(x+2)+3(x+2) = 0`
`<=> (2x+3)(x+2) = 0`
`<=>` \(\left[ \begin{array}{l}2x+3=0\\x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-3\\x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=-2\end{array} \right.\)
Vậy `S= {-(3)/2; -2}`
