I. Cho bt:
P1
$P= \frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}$
a, Rút gọn
b, Tìm MaxP
P2
$P=\left ( \frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}} \right ):\left ( 1+\frac{x+y+2xy}{1-xy} \right )$
a,Rút gọn
b,Tính giá trị của P khi $x=\frac{2}{2+\sqrt{3}}$
c,Tìm MaxP
b2. $P=\frac{(\sqrt{x}+\sqrt{y})(1+\sqrt{xy})+(\sqrt{x}-\sqrt{y})(1-\sqrt{xy})}{(1-\sqrt{xy})(1+\sqrt{xy})}:\frac{1-xy+x+y+2xy}{1-xy}=\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}.\frac{1-xy}{x+y+xy+1}=\frac{2\sqrt{x}(y+1)}{(x+1)(y+1)}=\frac{2\sqrt{x}}{x+1}$ (ĐK : $x,y\geq 0 xy\neq 1$ )
b)$P=\frac{2\sqrt{x}}{x+1},x=\frac{2}{2+\sqrt{3}}=\frac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=4-2\sqrt{3}=(\sqrt{3}-1)^2=>P=\frac{2(\sqrt{3}-1)}{5-2\sqrt{3}}$
c)$(\sqrt{x}-1)^2\geq 0 =>x+1\geq 2\sqrt{x}=>P=\frac{2\sqrt{x}}{x+1}\leq 1 .$
Dấu bằng xảy ra khi và chỉ khi x=1