Question

\(\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+2\right)\)
Giúp mình với ạ

Answer 1

\(\left(x-1\right)^2+\left(x-3\right)^2=2\left(x-2\right)\left(x+2\right)\)

\(x^2-2x+1+x^2+6x+9=2x^2-8\)

\(2x^2+4x+10=2x^2-8\)

\(4x=-18\)

x=\(\frac{-9}{2}\)

Answer 2

Ta có: \(\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2\left(x^2-4\right)\)

\(\Leftrightarrow2x^2+4x+10=2x^2-8\)

\(\Leftrightarrow2x^2+4x+10-2x^2+8=0\)

\(\Leftrightarrow4x+18=0\)

\(\Leftrightarrow4x=-18\)

hay \(x=-\frac{9}{2}\)

Vậy: \(x=-\frac{9}{2}\)