Question

Giúp mình với ạ

Answer 1

ĐKXĐ: x>=-3/2

Ta có: \(x^2-3=2\sqrt{2x+3}\)

=>\(x^2-3-6=2\sqrt{2x+3}-6\)

=>\(x^2-9=2\left(\sqrt{2x+3}-3\right)\)

=>\(\left(x-3\right)\left(x+3\right)=2\cdot\frac{2x+3-9}{\sqrt{2x+3}+3}\)

=>\(\left(x-3\right)\left(x+3\right)=2\cdot\frac{2x-6}{\sqrt{2x+3}+3}\)

=>\(\left(x-3\right)\left(x+3\right)-\frac{4\left(x-3\right)}{\sqrt{2x+3}+3}=0\)

=>\(\left(x-3\right)\left(x+3-\frac{4}{\sqrt{2x+3}+3}\right)=0\)

=>x-3=0

=>x=3(nhận)