~ Bạn tham khảo ~
`a, x^2 - 4x + 4 = 9(x-2)`
`<=> (x-2)^2 - 9(x-2)= 0`
`<=> (x-2)(x-2-9)=0`
`<=> (x-2)(x-11)=0`
`<=> [(x-2=0),(x-11=0):}`
`<=> [(x=2),(x=11):}`
Vậy `x={2;11}`
`b, 4x^2 + 12x + 9 = (5-x)^2`
`<=> (2x+3)^2 - (5-x)^2 = 0`
`<=> (2x+3-5+x)(2x+3+5-x) = 0`
`<=> (3x-2)(x+8)=0`
\(⇔\left[ \begin{array}{l}3x-2=0\\x+8=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac23\\x=-8\end{array} \right.\)
Vậy `x={2/3;-8}`
