\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
\(B=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x-3}\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b,để B=-3\(< =>\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=-3\)\(=>\sqrt{x}+1=-3\sqrt{x}+9\)
\(< =>4\sqrt{x}=8< =>x=4\left(loai\right)\)
vậy không có giá trị nào của x thỏa mãn để B=-3
c, \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
để B nguyên <=>\(\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
*\(\sqrt{x}-3=1=>x=16\left(TM\right)\)
*\(\sqrt{x}-3=-1=>x=4\left(loai\right)\)
*\(\sqrt{x}-3=2=>x=25\left(TM\right)\)
*\(\sqrt{x}-3=-2=>x=1\left(TM\right)\)
*\(\sqrt{x}-3=4=>x=49\left(TM\right)\)
*\(\sqrt{x}-3=-4=>\sqrt{x}=-1\)(vô lí)
vậy \(x\in\left\{1;49;16;25;\right\}\)
