Đáp án:a)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b)\(\left \{ {{x=\frac{1}{4}} \atop {x=4}} \right.\)
Giải thích các bước giải:
B=\(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}=\frac{x-2}{\sqrt{x}(\sqrt{x}+2)}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}-2}{\sqrt{x}(\sqrt{x}+2)}=\frac{(\sqrt{x}-1)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}\)
=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)
P=A·B=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}·\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b)\(2P=2\sqrt{x}+5 ⇒ 2\frac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+5\)
⇒ \(2\sqrt{x}+2=2x+5\sqrt{x}⇒ 2x+3\sqrt{x}-2=0 ⇒ 2x+4\sqrt{x}-\sqrt{x}-2=0\)
⇒ \((2\sqrt{x}-1)(\sqrt{x}+2)=0 ⇒\left \{ {{2\sqrt{x}-1=0} \atop {\sqrt{x}+2=0}} \right.\)
⇒ \(\left \{ {{\sqrt{x}=\frac{1}{2}} \atop {\sqrt{x}=-2}} \right.⇒\left \{ {{x=\frac{1}{4}} \atop {x=4}} \right.\)
