a: KHi x=64 thì \(A=\dfrac{8-1}{8}=\dfrac{7}{8}\)
b: \(P=A-B=\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
`a)` Thay `x=64` (t/m đk) vào `A` có:
`A=[\sqrt{64}-1]/\sqrt{64}=7/8`
`b)` Với `x > 0,x \ne 1` có:
`A-B=[\sqrt{x}-1]/\sqrt{x}-[\sqrt{x}-1]/[x-1]`
`=[(\sqrt{x}-1)(x-1)-\sqrt{x}(\sqrt{x}-1)]/[\sqrt{x}(x-1)]`
`=[x\sqrt{x}-\sqrt{x}-x+1-x+\sqrt{x}]/[\sqrt{x}(x-1)]`
`=[x\sqrt{x}-2x+1]/[\sqrt{x}(x-1)]`
`=[x\sqrt{x}-x-x+\sqrt{x}-\sqrt{x}+1]/[\sqrt{x}(x-1)]`
`=[(\sqrt{x}-1)(x-\sqrt{x}-1)]/[\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)]`
`=[x-\sqrt{x}-1]/[\sqrt{x}(\sqrt{x}+1)]`
a, Khi x = 64: \(A=\dfrac{\sqrt{64}-1}{\sqrt{64}}=\dfrac{8-1}{8}=\dfrac{7}{8}\)
b, \(A-B=\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
