Đáp án `+` Giải thích các bước giải `!`
`a)`
`(x-2)^2-(x^2-3x) = 9`
`<=> x^2-4x+4-x^2+3x = 9`
`<=> (x^2-x^2)+(-4x+3x)+4 = 9`
`<=> -x = 5`
`<=> x = -5`
Vậy `S= {-5}`
`b)`
`(5x-2)^2 = (4-x)^2`
`<=> (5x-2)^2-(4-x)^2 = 0`
`<=> (5x-2-4+x)(5x-2+4-x) = 0`
`<=> (6x-6)(4x+2) = 0`
`<=> 6(x-1). 2(2x+1) = 0`
`<=> 12(x-1)(2x+1) = 0`
`<=> (x-1)(2x+1) = 0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\2x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\2x=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S= {1; -(1)/2}`
`c)`
`x^2-4x-5 = 0`
`<=> x^2-5x+x-5 = 0`
`<=> (x^2-5x)+(x-5) = 0`
`<=> x(x-5)+(x-5) = 0`
`<=> (x+1)(x-5) = 0`
`⇔` \(\left[ \begin{array}{l}x+1=0\\x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)
Vậy `S= {-1; 5}`
