Question

Giải phương trình sau

a) (x-2)^2+2(x-4)=(x-4)(x-2)

b) (x-1)(2x-3)-3(x-2)=2(x-1)^2

Answer 1

$(x-2)^2+2(x-4)=(x-4)(x-2)$

\(\Leftrightarrow x^2-4x+4+2x-8=x^2-6x+8\)

\(\Leftrightarrow4x-12=0\Leftrightarrow4x=12\Leftrightarrow x=3\)

Vậy \(x=3\)

Answer 2

$(x-2)^2+2(x-4)=(x-4)(x-2)$

\(\Leftrightarrow x^2-4x+4+2x-8=x^2-6x+8 \\\ \Leftrightarrow 2x-12=0 \\\ \Leftrightarrow x=6\)

Vậy $x=6$