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