Bài 2:
`a)` Với `x>=0; x\ne1` ta có:
`Q=(\sqrt{x}/(1-\sqrt{x})+\sqrt{x}/(1+\sqrt{x}))+(3-\sqrt{x})/(x-1)`
`Q=(\sqrt{x}(1+\sqrt{x}+\sqrt{x}(1-\sqrt{x})/((1-\sqrt{x})(1+\sqrt{x}))-(3-\sqrt{x})/(1-x)`
`Q=(\sqrt{x}+x+\sqrt{x}-x-3+\sqrt{x})/((1-\sqrt{x})(1+\sqrt{x}))`
`Q=(3\sqrt{x}-3)/((1-\sqrt{x})(1+\sqrt{x}))`
`Q=(3(\sqrt{x}-1))/((1-\sqrt{x})(1+\sqrt{x}))`
`Q=-3/(\sqrt{x}+1)`
Vậy `Q=-3/(\sqrt{x}+1)` với `x>=0;x\ne1`
`b)` Với `x>=0;x\ne1` thì
`Q=-1 -> -3/(\sqrt{x}+1)=-1`
`<=> \sqrt{x}+1=3`
`<=> \sqrt{x}=2`
`<=> x=4` (tm)
Vậy `x=4` thì `Q=-1`
Bài 3:
`A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}`
`A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3+2.2.\sqrt{3}+4}}}}`
`A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(\sqrt{3}+2)}}}`
`A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}`
`A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2.5.\sqrt{3}+3}}}`
`A=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}`
`A=\sqrt{4+\sqrt{25}}`
`A=\sqrt{4+5}`
`A=\sqrt{9}`
`A=3`
Vậy `A=3`
