Question

Giải phưong trình:

\(\sqrt{2x^2+6x-8}+\sqrt{2x^2+4x-6}-3\sqrt{x+4}=3\sqrt{x+3}+1\)

hộ e nha

Answer 1

ĐK : \(x\ge1\)

\(\Leftrightarrow...\sqrt{\left(x-1\right)\left(x+4\right)}+\sqrt{\left(x-1\right)\left(x+3\right)}-3\sqrt{x+4}-3\sqrt{x+3}=1\)

\(\Leftrightarrow\sqrt{x-1}\cdot\sqrt{x-4}+\sqrt{x-1}\cdot\sqrt{x+3}-3\cdot\left(\sqrt{x+4}+\sqrt{x+3}\right)=1\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+4}+\sqrt{x+3}\right)-3\left(\sqrt{x+4}+\sqrt{x+3}=1\right)\)

\(\Leftrightarrow\left(\sqrt{x-1}-3\right)\left(\sqrt{x+4}+\sqrt{x+3}\right)=1\)

Làm nốt nhé :)