Giải thích các bước giải:
a.Hàm số xác định
$\to\begin{cases} x\ge 0\\ x-1\ne 0\\ x+2\sqrt{x}+1\ne 0\end{cases}$
$\to\begin{cases} x\ge 0\\ x\ne 1\end{cases}$
Ta có:
$P=(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1})\cdot \dfrac{1-x}{2}$
$\to P=(\dfrac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2})\cdot \dfrac{1-x}{2}$
$\to P=(\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)^2})\cdot \dfrac{1-x}{2}$
$\to P=\dfrac{-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)^2}\cdot \dfrac{(1-\sqrt{x})(1+\sqrt{x})}{2}$
$\to P=\dfrac{-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)^2}\cdot \dfrac{-(\sqrt{x}-1)(1+\sqrt{x})}{2}$
$\to P=\dfrac{\sqrt{x}}{\sqrt{x}+1}$
b.Ta có:
$x=7-4\sqrt{3}$
$\to x=4-4\sqrt{3}+3$
$\to x=(2-\sqrt{3})^2$
$\to \sqrt{x}=2-\sqrt{3}$
$\to P=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}$
$\to P=\dfrac{3-\sqrt{3}}{6}$
c.Ta có:
$P=\dfrac{\sqrt{x}}{\sqrt{x}+1}$
$\to P\ge 0$
$\to GTNN_P=0$ khi đó $x=0$
Không tồn tại $GTLN_P$
