Question

Answer 1

1) Tính $A$ khi $x=64$

$A=\dfrac{2}{\sqrt{64}-2}$

$=\dfrac{2}{8-2}$

$=\dfrac{2}{6}$

$=\dfrac{1}{3}$

2) Chứng minh $B=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}$

$4-x=(2-\sqrt{x})(2+\sqrt{x})=-(\sqrt{x}-2)(\sqrt{x}+2)$

$B=\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\dfrac{3(\sqrt{x}+2)+(\sqrt{x}+1)(\sqrt{x}-2)+2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\dfrac{3\sqrt{x}+6+x-2\sqrt{x}+\sqrt{x}-2+2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\dfrac{x+4\sqrt{x}+4}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\dfrac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}$

3) Cho $P=\dfrac{A}{B}$. Tìm $x$ để $P\ge\dfrac{2}{x+2}$

$P=\dfrac{2}{\sqrt{x}-2}:\dfrac{\sqrt{x}+2}{\sqrt{x}-2}$

$=\dfrac{2}{\sqrt{x}+2}$

$\dfrac{2}{\sqrt{x}+2}\ge\dfrac{2}{x+2}$

$\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{x+2}$

$x+2\ge\sqrt{x}+2$

$x\ge\sqrt{x}$

$x\ge0$

Kết hợp điều kiện: $x\ge0,\ x\ne4$