các bn giúp mik vs ạ

Đáp án:

`a)`

`(x-1)^2=4`

`=>(x-1)^2-4=0`

`=>(x-1)^2-2^2=0`

`=>(x-1-2)(x-1+2)=0`

`=>(x-3)(x+1)=0`

`=>` \(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\) 

Vậy `x\in{-1;3}`

`b)`

`(2-x)^2=9`

`=>(2-x)^2-9=0`

`=>(2-x)^2-3^2=0`

`=>(2-x-3)(2-x+3)=0`

`=>(-x-1)(5-x)=0`

`=>-(x+1)(5-x)=0`

`=>` \(\left[ \begin{array}{l}x+1=0\\5-x=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\) 

Vậy `x\in{-1;5}`

`c)`

`(2x+1)^2=25`

`=>(2x+1)^2-25=0`

`=>(2x+1)^2-5^2=0`

`=>(2x+1-5)(2x+1+5)=0`

`=>(2x-4)(2x+6)=0`

`=>2(x-2).2(x+3)=0`

`=>4(x-2)(x+3)=0`

`=>` \(\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

Vậy `x\in{-3;2}`

--Áp dụng HĐT: `A^2-B^2=(A-B)(A+B)`