Đáp án:
Giải thích các bước giải:
a/ $P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-\sqrt{x}}$
ĐKXĐ: $x \geq 0$ và $x \neq 4$
$P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{(\sqrt{x}+2)(2-\sqrt{x})}$
$P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$P=\dfrac{(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}.(\sqrt{x}-2)-2-5\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$P=\dfrac{x+\sqrt{x}+2\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$P=\dfrac{3x-6\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$P=\dfrac{3\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}$
$P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}$
b/ Để $P=2$ thì:
$\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2$
⇔ $3\sqrt{x}=2\sqrt{x}+4$
⇔ $\sqrt{x}=4$
⇔ $x=16$
Vậy $x=16$ thì $P=2$
Chúc bạn học tốt !!
