`a) x^2+2x+1=9`
`<=> (x+1)^2=3^2`
`<=> (x+1)^2-3^2=0`
`<=> (x+1-3)(x+1+3)=0`
`<=> (x-2)(x+4)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\x+4=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
Vậy `x∈{2;-4}`
`b) (2x-5)^2=(x+2)^2`
`<=> (2x-5)^2-(x+2)^2=0`
`<=> (2x-5-x-2)(2x-5+x+2)=0`
`<=> (x-7)(3x-3)=0`
`<=> 3(x-7)(x-1)=0`
`<=>`\(\left[ \begin{array}{l}x-7=0\\x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=7\\x=1\end{array} \right.\)
Vậy `x∈{7;1}`
`c) (x+1)^2=4(x^2+2x+1)`
`<=> (x+1)^2=4(x+1)^2`
`<=> 4(x+1)^2-(x+1)^2=0`
`<=> 3(x+1)^2=0`
`<=> x+1=0`
`<=>x=-1`
Vậy `x=-1`
