Question

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Q =\(\left(\frac{1}{\sqrt{x}-4}+3\right).\frac{x-16}{3-\sqrt{x}-11}\)

làm nhanh giúp mình nha

Answer 1

\(ĐKXĐ:x\ne16\)

\(Q=\frac{1+3\sqrt{x}-12}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{3-\sqrt{x}-11}\)

\(=\frac{\left(3\sqrt{x}-11\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)

Answer 2

\(\left(\frac{1}{\sqrt{x}-4}+3\right).\frac{x-16}{3-\sqrt{x}-11}=\left(\frac{1}{\sqrt{x}-4}+\frac{3\left(\sqrt{x}-4\right)}{\sqrt{x}-4}\right).\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)

\(=\frac{1+3\left(\sqrt{x}-4\right)}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}=\frac{1+3\sqrt{x}-12}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)

\(=\frac{3\sqrt{x}-11}{\sqrt{x}-4}.\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{-\left(\sqrt{x}+8\right)}=\frac{\left(3\sqrt{x}-11\right)\left(\sqrt{x}+4\right)}{-\left(\sqrt{x}+8\right)}\)

Answer 3

\(ĐKXĐ:x\ne16\)

\(=\frac{1+2\sqrt{x}-12}{\sqrt{x-4}}\).\(\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}{3-\sqrt{x}-11}\)

\(=\frac{\left(3\sqrt{x}-11\right)\left(\sqrt{x}+4\right)}{-\sqrt{x}-8}\)