Question

Answer 1

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(A=\dfrac{x+4}{4-x}-\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2+\sqrt{x}}\)

\(=\dfrac{x+4+\sqrt{x}+2+x-2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(=\dfrac{2x-\sqrt{x}+6}{4-x}\)

Answer 2

a) ĐKXĐ: x ≥ 0, x ≠ 4

b) \(A=\dfrac{x+4}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

        \(=\dfrac{x+4-\sqrt{x}-2-x+2\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)

        \(=\dfrac{4+2\sqrt{x}}{\sqrt{x}+2}\)