a)\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)
=\(\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)\)=\(\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
b) ĐKXĐ:\(x\ge0,x\ne9,x\ne4\)
c)M=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)\(-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
M=\(\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
M=\(\dfrac{2-\sqrt{x}}{-\left(\sqrt{x}-3\right)\left(2-\sqrt{x}\right)}\)=\(\dfrac{-1}{\sqrt{x}-3}\)
d) để M<-1
\(\Rightarrow\dfrac{-1}{\sqrt{x}-3}< -1\Rightarrow-1>-1\left(\sqrt{x}-3\right)\)
\(\Rightarrow-1>3-\sqrt{x}\)\(\Rightarrow-\sqrt{x}< -4\Rightarrow x>16\)
e) để \(M\in Z\)\(\Rightarrow\)\(\sqrt{x}-3\inƯ\left(-1\right)=\left\{1;-1\right\}\)
| \(\sqrt{x}-3\) | 1 | -1 |
| x | 16(TM) | 4(KTM) |
vậy x=4 thì \(M\in Z\)
