Question

rút gọn

\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}+\frac{\sqrt{x}+8}{4-\sqrt{x}}\)

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Answer 1

ĐKXĐ: \(x\ge0;x\ne16\)

\(=\frac{x\sqrt{x}-2x+28}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}-4\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-4\right)}\)

\(=\frac{x\sqrt{x}-2x+28-x+8\sqrt{x}-16-x-9\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(=\frac{x\sqrt{x}-4x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(=\sqrt{x}-1\)