Đáp án+Giải thích các bước giải:
a,
`x=7-4\sqrt{3}`
`⇔x=(2-\sqrt{3})^2`
`A=\frac{2+\sqrt{x}}{\sqrt{x}}(x>0)`
Thay `x=(2-\sqrt{3})^2` vào `A` ta có:
`⇔A=\frac{2+\sqrt{(2-\sqrt{3})^2}}{(2-\sqrt{3})^2}`
`⇔A=\frac{2+2-\sqrt{3}}{2-\sqrt{3}}`
`⇔A=\frac{4-\sqrt{3}}{2-\sqrt{3}}`
`⇔A=\frac{5+2\sqrt{3}}{4-3}`
`⇔A=\frac{5+2\sqrt{3}}{1}`
`⇔A=5+2\sqrt{3}`
Vậy với `x=7-4\sqrt{3}` thì `A=5+2\sqrt{3}`
b,
`B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}(x>0)`
`B=\frac{(\sqrt{x}-1)(\sqrt{x}+1)+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}`
`B=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}`
`B=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+1)}`
`B=\frac{\sqrt{x}+2}{\sqrt{x}+1}`
c,
`\frac{A}{B}>\frac{3}{2}`
`⇔\frac{\frac{2+\sqrt{x}}{\sqrt{x}}}{\frac{\sqrt{x}+2}{\sqrt{x}+1}}>\frac{3}{2}`
`⇔\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}`
`⇔\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0`
`⇔\frac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0`
`⇔\frac{-\sqrt{x}+2}{2\sqrt{x}}>0`
`2\sqrt{x}>0∀x>0`
`⇒-\sqrt{x}+2>0`
`⇔\sqrt{x}<2`
`⇔x<4`
Kết hợp điều kiện xác định
Vậy `0<x<4` thì `\frac{A}{B}>\frac{3}{2}`
