Question

Tìm x :
a,x²-2x+1=4
b,16-(x-3)²=0

Answer 1

a, `x^2-2x+1=4`

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

`<=> [(x-1=2),(x-1=-2);}`

`<=> [(x=3),(x=-1):}`

b, `16-(x-3)^2=0`

`<=>(x-3)^2=4^2=(-4)^2`

`<=> [(x-3=4),(x-3=-4):}`

`<=> [(x=7),(x=-1):}`

 

Answer 2

a) Ta có: \(x^2-2x+1=4\)

\(\Leftrightarrow\left(x-1\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

b) Ta có: \(16-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)