a: \(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Ta có: \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{-3}{\sqrt{x}+3}\)
b: \(P<\frac12\)
=>\(P-\frac12<0\)
=>\(\frac{-3}{\sqrt{x}+3}-\frac12<0\)
=>\(\frac{-6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}<0\)
=>\(\frac{-\sqrt{x}-9}{2\left(\sqrt{x}+3\right)}\) <0(luôn đúng)
c: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ
=>\(\frac{3}{\sqrt{x}+3}\le1\forall x\) thỏa mãn ĐKXĐ
=>\(-\frac{3}{\sqrt{x}+3}\ge-1\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
Em cần giúp câu c và d ạ, mn giúp em với em đang cần gấp
