$M=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}}$ $(x\neq4;1;0;9)$
$=1-\dfrac{1}{\sqrt{x}-2}-1+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}}$
$=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-2}+\dfrac{x-2}{x-3\sqrt{x}}$
$=\dfrac{\sqrt{x}(\sqrt{x}-2)(x-3\sqrt{x})-(\sqrt{x}-1)(x-3\sqrt{x})+(x-2)(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
$=\dfrac{(x²-5x\sqrt{x}+6x)-(x\sqrt{x}-4x+3\sqrt{x})+(x²-3x\sqrt{x}+6\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
$=\dfrac{2x²-9x\sqrt{x}+10x+3\sqrt{x}-4}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
$=\dfrac{(\sqrt{x}-1)(2x\sqrt{x}-7x+3\sqrt{x}+6)+10}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
$=\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)(2x-3\sqrt{x}-3)+10}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
$=\dfrac{2x-3\sqrt{x}-3}{(x-3\sqrt{x})}+\dfrac{10}{(\sqrt{x}-2)(\sqrt{x}-1)(x-3\sqrt{x})}$
Mình chỉ rút gọn được đến thế, không tìm $x$ cho $M<1$ được
