Đáp án:
Giải thích các bước giải:
Đặt $\sqrt{x}=y$, ta được:
$M=\frac{2y-9}{y^2-5y+6}+\frac{2y+1}{y-3}+\frac{y+3}{2-y}\\\Leftrightarrow M=\frac{2y-9}{(y-3)(y-2)}+\frac{2y+1}{y-3}-\frac{y+3}{y-2}\\\Leftrightarrow (y-3)(y-2)M=2y-9+(y-2)(2y+1)-(y+3)(y-3)\\\Leftrightarrow (y-3)(y-2)M=(2y-9)+(2y^2-3y-2)-(y^2-9)\\\Leftrightarrow (y-3)(y-2)M=y^2-y-2\\\Leftrightarrow M=\frac{(y-2)(y+1)}{(y-3)(y-2)}\\\Leftrightarrow M=\frac{y+1}{y-3}\\\Leftrightarrow M=\frac{\sqrt{x}+1}{\sqrt{x}-3}$
