Question

giúp em với ạ

Answer 1

Ta có: \(M=\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{6+\sqrt{x}}{x-4}\)

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

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

Ta có: \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{3}{\sqrt{x}+2}\le\frac32\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{-3}{\sqrt{x}+2}\ge\frac{-3}{2}\forall x\) thỏa mãn ĐKXĐ

=>Chọn C