Question

giải phương trình vô tỉ sau 

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

Answer 1

\(\sqrt{\left(x+2\right)\left(x-1\right)}+3\sqrt{x+2}=2\left(x+2\right)\)(đk bn tự xd nhé)

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-1}+3-2\sqrt{x+2}\right)\)=0

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\\sqrt{x-1}+3=2\sqrt{x+2}\left(1\right)\end{cases}}\)

giai (1) bn se co x=2 kl x=+-2