Question

\(2x^2-5x=\sqrt{x+1}+\sqrt{x-2}\) giải phương trình

Answer 1

ĐKXĐ: x>=2

Ta có: \(2x^2-5x=\sqrt{x+1}+\sqrt{x-2}\)

=>\(2x^2-5x-3=\sqrt{x+1}-2+\sqrt{x-2}-1\)

=>\(\left(x-3\right)\left(2x+1\right)=\frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x-2}+1}\)

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

=>x-3=0

=>x=3(nhận)