Đáp án:
1. $\dfrac45$
2. $ B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$
3. $ 0\le x\le 4, x\ne 1$
Giải thích các bước giải:
1. Khi $x=36\to \sqrt{x}=6$
$\to A=\dfrac{6-2}{6-1}=\dfrac45$
2.Ta có:
$B=\dfrac{x-5}{x-1}-\dfrac{2}{\sqrt{x}+1}-\dfrac{4}{1-\sqrt{x}}$
$\to B=\dfrac{x-5}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{4(\sqrt{x}+1)}{(\sqrt{x}+1)(1-\sqrt{x})}$
$\to B=\dfrac{x-5}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{2\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{4\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{x-5-(2\sqrt{x}-2)+(4\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{x+2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{(\sqrt{x}+1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$
3.Ta có:
$P=\dfrac AB$
$\to P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$
$\to P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
$\to P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}$
Để $\sqrt{P}<\dfrac12$
$\to 0\le P<\dfrac14$
Giải $0\le P$
$\to \dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge 0$
$\to \sqrt{x}-2\ge 0$
$\to \sqrt{x}\ge 2$
$\to x\ge 4(1)$
Giải $P<\dfrac14$
$\to \dfrac{\sqrt{x}-2}{\sqrt{x}+1}<\dfrac14$
$\to 4(\sqrt{x}-2)<\sqrt{x}+1$ vì $\sqrt{x}+1>0$
$\to 4\sqrt{x}-8<\sqrt{x}+1$
$\to 3\sqrt x<9$
$\to \sqrt x<3$
$\to 0\le x<9(2)$
Từ $(1), (2)$
$\to 0\le x\le 4$
Kết hợp đkxđ $\to 0\le x\le 4, x\ne 1$
