`a)` Thay `x=9` vào `A` có:
`A=[9+\sqrt{9}+1]/[\sqrt{9}+2]=13/5`
`b)` Với `x > 0,x \ne 1` có:
`B=[2\sqrt{x}]/[\sqrt{x}-1]-[x-\sqrt{x}+2]/[x-\sqrt{x}]`
`B=[2x-x+\sqrt{x}-2]/[\sqrt{x}(\sqrt{x}-1)]`
`B=[x+\sqrt{x}-2]/[\sqrt{x}(\sqrt{x}-1)]`
`B=[(\sqrt{x}-1)(\sqrt{x}+2)]/[\sqrt{x}(\sqrt{x}-1)]`
`B=[\sqrt{x}+2]/\sqrt{x}`
a.Thế \(x=9\) vào `A`, ta được:
\(A=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{9+\sqrt{9}+1}{\sqrt{9}+2}=\dfrac{13}{5}\)
b.\(B=\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}}\)
\(B=\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{x-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{2x-x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
