`***`lời giải`***`
Bài 1.
`A=3\sqrt{49}-\sqrt{25} `
`=3.7-5`
`=21-5=16`
`B=\sqrt{(3-2\sqrt{5})^2}-\sqrt{20}`
`=|3-2\sqrt{5}|-2\sqrt{5}`
`=2\sqrt{5}-3-2\sqrt{5}`
`=-3`
Bài 2.
ĐKXĐ:`x>0;xne1`
a)
`P=(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-\sqrt{x}}):(\frac{\sqrt{x}+1}{3})`
`=(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}):(\frac{\sqrt{x}+1}{3})`
`=\frac{x+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}:\frac{\sqrt{x}+1}{3}`
`=\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}.\frac{3}{\sqrt{x}+1}`
`=\frac{3}{\sqrt{x}-1}`
b)
Ta có: `P=1`
`=>\frac{3}{\sqrt{x}-1}=1`
`<=>\frac{3}{\sqrt{x}-1}-1=0`
`<=>\frac{3-(\sqrt{x}-1)}{\sqrt{x}-1}=0`
`<=>\frac{3-\sqrt{x}+1}{\sqrt{x}-1}=0`
`<=>\frac{4-\sqrt{x}}{\sqrt{x}-1}=0`
`=>4-\sqrt{x}=0`
`<=>-\sqrt{x}=-4`
`<=>\sqrt{x}=4`
`<=>x=16(TMĐK)`
Vậy `x=16` để `P=1`
