a) x^2 + 2x + 1 = 9 b) (2x – 5)^2 = (x + 2)^2 c) (x + 1)^2 = 4(x^2 + 2x + 1)

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1. Đáp án+Giải thích các bước giải:

   `a,x^2+2x+1=9`

   `=>(x+1)^2=9`

   `=>(x-2)(x+4)=0`

   `=>x=2\or\x=-4`

   `b,(2x-5)^2=(x+2)^2`

   `=>(x-7)(3x-3)=0`

   `=>(x-7)(x-1)=0`

   `=>x=7\or\x=1`

   `c,(x+1)^2=4(x^2+2x+1)`

   `=>(x+1)^2=4(x+1)^2`

   `=>3(x+1)^2=0`

   `=>x+1=0`

   `=>x=-1`

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2. `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`

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