ĐKXĐ : \(x\ge0\)
Ta có :
\(P=\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+2-x+\sqrt{x}-1-x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{2-\sqrt{x}}{x-\sqrt{x}+1}\)
Vậy...
b/ Ta có :
\(P\le\frac{4}{3}\)
\(\Leftrightarrow\frac{2-\sqrt{x}}{x-\sqrt{x}+1}\le\frac{4}{3}\)
\(\Leftrightarrow\frac{6-3\sqrt{x}}{3.\left(x-\sqrt{x}+1\right)}-\frac{4x-4\sqrt{x}+4}{3.\left(x-\sqrt{x}+1\right)}\le0\)
\(\Leftrightarrow\frac{7\sqrt{x}-4x+2}{3\left(x-\sqrt{x}+1\right)}\le0\)
Mà \(3\left(x-\sqrt{x}+1\right)\ge0\forall x\)
\(\Leftrightarrow-\left(4x-7\sqrt{x}-2\right)\le0\)
Tự giải tiếp nhé ~
