Câu 9:
a: \(P=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\dfrac{1-x}{x}\)
b: Để P>1/2 thì \(P-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{1-x}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow2\left(1-x\right)-x>0\)
=>-3x+2>0
=>-3x>-2
hay x<2/3
Vậy: 0<x<2/3
Câu `9:`
`a)` Với `x > 0` có:
`P=(1/[x+\sqrt{x}]-1/[\sqrt{x}+1]):\sqrt{x}/[x+2\sqrt{x}+1]`
`P=[1-\sqrt{x}]/[\sqrt{x}(\sqrt{x}+1] .[(\sqrt{x}+1)^2]/\sqrt{x}`
`P=[1-x]/x`
`b)` Với `x > 0` có:
`P > 1/2<=>[1-x]/x > 1/2`
`<=>[1-x]/x-1/2 > 0`
`<=>[2(1-x)-x]/[2x] > 0`
Mà `x > 0<=>2x > 0`
`=>2-2x-x > 0<=>-3x > -2<=>x < 2/3`
Kết hợp `x > 0`
`=>0 < x < 2/3`
Câu `10`:
`a)` Với `a > 0,a \ne 9` có:
`P=(1/[\sqrt{a}-3]+1/[\sqrt{a}+3])(1-3/\sqrt{a})`
`P=[\sqrt{a}+3+\sqrt{a}-3]/[(\sqrt{a}-3)(\sqrt{a}+3)] . [\sqrt{a}-3]/\sqrt{a}`
`P=[2\sqrt{a}]/[(\sqrt{a}-3)(\sqrt{a}+3)] . [\sqrt{a}-3]/\sqrt{a}`
`P=2/[\sqrt{a}+3]`
`b)` Với `a > 0,a \ne 9` có:
`P > 1/2<=>2/[\sqrt{a}+3] > 1/2`
`<=>2/[\sqrt{a}+3]-1/2 > 0`
`<=>[4-\sqrt{a}-3]/[2(\sqrt{a}+3)] > 0`
Mà `a > 0=>2(\sqt{a}+3) > 0`
`=>1-\sqrt{a} > 0<=>\sqrt{a} < 1<=>a < 1`
Kết hợp `a > 0,a \ne 9`
`=>0 < a < 1`
