Đáp án+Giải thích các bước giải:
`1)`
`a)x≥0;x\ne9`
`M=(2\sqrt{x})/(\sqrt{x}+3)+(\sqrt{x}+1)/(\sqrt{x}-3)+(3-11\sqrt{x})/(9-x)`
`=(2\sqrt{x})/(\sqrt{x}+3)+(\sqrt{x}+1)/(\sqrt{x}-3)-(3-11\sqrt{x})/(x-9)`
`=((2\sqrt{x}).(\sqrt{x}-3))/((\sqrt{x}+3).(\sqrt{x}-3))+((\sqrt{x}+1).(\sqrt{x}+3))/((\sqrt{x}-3).(\sqrt{x}+3))-(3-11\sqrt{x})/((\sqrt{x}-3).(\sqrt{x}+3)`
`=(2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x})/((\sqrt{x}-3).(\sqrt{x}+3)`
`=(3x+9\sqrt{x})/((\sqrt{x}-3).(\sqrt{x}+3)`
`=(3\sqrt{x}.(\sqrt{x}+3))/((\sqrt{x}-3).(\sqrt{x}+3)`
`=(3\sqrt{x})/(\sqrt{x}-3)`
Vậy `M=(3\sqrt{x})/(\sqrt{x}-3)` với `x≥0;x\ne9`
`b)`
`x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}(tmđk)`
`=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}`
`=\sqrt{\sqrt{3}-\sqrt{(\sqrt{3}-1)^2}`
`=\sqrt{\sqrt{3}-|\sqrt{3}-1|}`
`=\sqrt{\sqrt{3}-\sqrt{3}+1}(Vì:\sqrt{3}-1>0)`
`=\sqrt{1}=1`
Thay `x=1` vào `M` ta có:
`(3\sqrt{1})/(\sqrt{1}-3)=(3)/(-2)=-(3)/(2)`
Vậy `M=(-3)/(2)` khi `x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}`
`2)`
`a)x>0;x\ne1`
`P=((\sqrt{x})/(\sqrt{x}-1)+(\sqrt{x})/(x-1)):((2)/(x)-(2-x)/(x.(\sqrt{x}+1)))`
`=((\sqrt{x}.(\sqrt{x}+1))/((\sqrt{x}-1).(\sqrt{x}+1))+(\sqrt{x})/((\sqrt{x}-1).(\sqrt{x}+1))):((2.(\sqrt{x}+1))/(x.(\sqrt{x}+1))-(2-x)/(x.(\sqrt{x}+1)))`
`=(x+\sqrt{x}+\sqrt{x})/((\sqrt{x}-1).(\sqrt{x}+1)):(2\sqrt{x}+2-2+x)/(x.(\sqrt{x}+1))`
`=(x+2\sqrt{x})/((\sqrt{x}-1).(\sqrt{x}+1)).(x.(\sqrt{x}+1))/(x+2\sqrt{x})`
`=(x)/(\sqrt{x}-1)`
Vậy `P=(x)/(\sqrt{x}-1)` với `x>0;x\ne1`
`b)`
`x=(2)/(2-\sqrt{3})-2\sqrt{3}(tmđk)`
`=(2.(2+\sqrt{3}))/((2-\sqrt{3}).(2+\sqrt{3}))-2\sqrt{3}`
`=(4+2\sqrt{3})/(4-3)-2\sqrt{3}`
`=4+2\sqrt{3}-2\sqrt{3}=4`
Thay `x=4` vào `P` ta có:
`(4)/(\sqrt{4}-1)=(4)/(2-1)=4`
Vậy `P=4` khi `x=(2)/(2-\sqrt{3})-2\sqrt{3}`
