$a)\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} }$
với `x>=0;x#4;x#9`
`=`$\dfrac{2\sqrt[]{x}-9}{(\sqrt[]{x}-2)( \sqrt[]{x}-3) }-$$\dfrac{(\sqrt[]{x}-3)(\sqrt[]{x} +3)}{(\sqrt[]{x}-2)( \sqrt[]{x}-3)}+$$\dfrac{(2\sqrt[]{x}+1)(\sqrt[]{x}-2) }{ (\sqrt[]{x}-2)( \sqrt[]{x}-3)}$
`=`$\dfrac{2\sqrt[]{x}-9-x+9+2x-4\sqrt[]{x}+ \sqrt{x}-2}{(\sqrt[]{x}-2)( \sqrt[]{x}-3) }$
`=`$\dfrac{x-\sqrt[]{x}-2}{(\sqrt[]{x}-2)(\sqrt[]{x} -3) }$
`=`$\dfrac{(\sqrt[]{x}+1)(\sqrt[]{x}-2)}{(\sqrt[]{x}-2)(\sqrt[]{x} -3) }$
`=`$\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3 }$
`b)` ta có:
$11-6\sqrt[]{2}$ `=3^2-`$6\sqrt[]{2}+2$ `=(3-`$\sqrt[]{2})^2$
thay $x=(3-\sqrt[]{2})^2(tm)$ vào` M` ta có:
$\dfrac{\sqrt[]{(3-\sqrt[]{2})^2}+1}{\sqrt[]{ (3-\sqrt[]{2})^2}-3 }$
`=`$\dfrac{|3-\sqrt[]{2}|+1}{|3-\sqrt[]{2}|-3 }$
`=`$\dfrac{3-\sqrt[]{2}+1}{3-\sqrt[]{2}-3 }$ `=`$\dfrac{4-\sqrt[]{2}}{-\sqrt[]{2} }$
`=`$\dfrac{\sqrt[]{2}(2\sqrt[]{2}-1)}{-\sqrt[]{2}}$ `=`$-2\sqrt[]{2}+1$
`c)`$\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3 }=2$
`<=>`$\sqrt[]{x}+1=2$ $(\sqrt[]{x}-3)$
`<=>`$\sqrt[]{x}+1=2$ $\sqrt[]{x}-6$
`<=>`$-\sqrt[]{x}=-7$
`<=>x=49`
`d)xét:M=`$\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3 }$
ta thấy $\sqrt[]{x}+1$`>0∀x>=0;`
`=>M<1<=>`$\sqrt[]{x}-3<0$
`<=>`$\sqrt[]{x}<3$
`<=>x<9`
kết hợp `đkxđ =>9>x>=0`
`e)`ta có:$\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}-3 }=$$1+\dfrac{4}{\sqrt[]{x}-3 }$
mà `1∈Z=>`$\dfrac{4}{\sqrt[]{x}-3 }∈Z$
`=>`$\sqrt[]{x}$` -3∈Ư(4)={1;-1;4;-4}`
`=>`$\sqrt[]{x}$ `∈{4;2;7;-1}`
`=>x∈{16;4;49;1}`
