Đáp án+Giải thích các bước giải:
Bài `4:`
`a)ĐKXĐ:a>=0;a\ne1;a\ne4`
`P=(\frac{a-\sqrt{a}}{\sqrt{a}-1}+1):(\frac{a+2\sqrt{a}}{\sqrt{a}+2}-1)`
`=(\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}+1):(\frac{\sqrt{a}(\sqrt{a}+2)}{\sqrt{a}+2}-1)`
`=(\sqrt{a}+1):(\sqrt{a}-1)`
`=\frac{\sqrt{a}+1}{\sqrt{a}-1}`
`b)P=4`
`<=>\frac{\sqrt{a}+1}{\sqrt{a}-1}=5`
`<=>\sqrt{a}+1=5(\sqrt{a}-1)`
`<=>\sqrt{a}+1=5\sqrt{a}-5`
`<=>5\sqrt{a}-\sqrt{a}=1+5`
`<=>4\sqrt{a}=6`
`<=>\sqrt{a}=3/2`
`<=>a=9/4(tm)`
Vậy `a=9/4` thì `P=5`
`c)a=3+2\sqrt{2}`
`->\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1`
Thế `\sqrt{a}=\sqrt{2}+1` vào `P`,ta có:
`P=\frac{\sqrt{2}+1+1}{\sqrt{2}+1-1}`
`=\frac{\sqrt{2}+2}{\sqrt{2}}`
`=\frac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}}`
`=\sqrt{2}+1`
Vậy nếu `x=\sqrt{2}+1` thì `P=\sqrt{2}+1`
`d)P=\frac{\sqrt{a}+1}{\sqrt{a}-1}`
`=1+\frac{2}{\sqrt{a}-1}`
`->2\vdots\sqrt{a}-1`
`->\sqrt{a}-1\inƯ(2)\in{+-1;+-2}`
mà `\sqrt{a}-1>=-1`
`->\sqrt{a}-1\in{+-1;2}`
`->a\in{0;4;9}`
Vậy với `a\in{0;4;9}` thì `P\inZZ`
`e)P<1`
`->\frac{\sqrt{a}+1}{\sqrt{a}-1}<1`
`->\frac{\sqrt{a}+1-\sqrt{a}+1]{\sqrt{a}-1}<0`
`->\frac{2}{\sqrt{a}-1}<0`
mà `2>0`
`=>\sqrt{a}-1<0`
`->a<1`
Vậy `0<=a<1` thì `P<1`
