`a, Đk: x ne 1, x >=0`.
`a, A = (x+sqrt x)/(x-1) + (3 sqrt x - 3)/(x-1) + (4 - 6 sqrt x)/(x-1)`
`<=> (x - 2 sqrt x + 1)/(x-1)`
`<=> (sqrt x - 1)^2/(x-1)`
`<=> (sqrt x - 1)/(sqrt x + 1)`
`b, A < 1/2 => (sqrt x-1)/(sqrt x +1) < 1/2`
`<=> sqrt x + 1 < 2 (sqrt x-1)`
`<=> sqrt x + 1 < 2 sqrt x - 2`
`<=> 3 < sqrt x`
`=> x > 9`.
`d, A = (sqrt x -1)/(sqrt x + 1) = (sqrt x + 1)/(sqrt x + 1) - 2/(sqrt x + 1) = 1 - 2/(sqrt x + 1)`.
Vì `sqrt x >=0 => sqrt x + 1 >=1 => A >= 1 - 2 = -1`
Dấu bằng xảy ra `<=> x = 0`.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+1\ne0\\\sqrt{x}-1\ne0\\1-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a) \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}+\dfrac{6\sqrt{x}-4}{1-x}\)
\(A=\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Đề thiếu. Tính giá trị của A tại x bằng mấy?
