Question

giúp e vs ạ

Answer 1

a: ĐKXĐ: x>=0; x<>1

\(P=\dfrac{\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}+1}{2\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

b: Để P=3/5 thì \(\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{3}{5}\)

=>\(5\sqrt{x}=3\sqrt{x}+6\)

=>\(2\sqrt{x}=6\)

=>\(\sqrt{x}=3\)

=>x=9(nhận)