Đáp án:
a.$P=x-1$
b.$ x>1 , x\ne 4$
Giải thích các bước giải:
a.ĐKXĐ: $x\ge 0, x\ne 1, 4$
Ta có:
$P=(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{x^2-3x}{x-\sqrt{x}-2}):(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-3\sqrt{x}+2})$
$\to P=(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-2)}+\dfrac{x^2-3x}{(\sqrt{x}-2)(\sqrt{x}+1)}):(\dfrac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}-2)}+\dfrac{1}{(\sqrt{x}-2)(\sqrt{x}-1)})$
$\to P=\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)+x^2-3x}{(\sqrt{x}-2)(\sqrt{x}+1)}:\dfrac{\sqrt{x}(\sqrt{x}-2)+1}{(\sqrt{x}-2)(\sqrt{x}-1)}$
$\to P=\dfrac{x-1+x^2-3x}{(\sqrt{x}-2)(\sqrt{x}+1)}:\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}-1)}$
$\to P=\dfrac{x^2-2x+1}{(\sqrt{x}-2)(\sqrt{x}+1)}:\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}-1)}$
$\to P=\dfrac{(x-1)^2}{(\sqrt{x}-2)(\sqrt{x}+1)}:\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}-2)(\sqrt{x}-1)}$
$\to P=\dfrac{(x-1)^2}{(\sqrt{x}-2)(\sqrt{x}+1)}\cdot\dfrac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-1)^2}$
$\to P=\dfrac{(\sqrt{x}-1)^2(\sqrt{x}+1)^2}{(\sqrt{x}-2)(\sqrt{x}+1)}\cdot\dfrac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-1)^2}$
$\to P=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)$
$\to P=x-1$
b.Để $P>0$
$\to x-1>0$
$\to x>1$
Kết hợp $đkxđ \to x>1 , x\ne 4$
