Đáp án:a)$-9\sqrt{3}$
b)2
c)$3(-1+\sqrt{6})$
Giải thích các bước giải:
a)$\sqrt{12}+\sqrt{27}-\sqrt{108}-\sqrt{192}$
=$\sqrt{4·3}+\sqrt{9·3}-\sqrt{36·3}-\sqrt{64·3}$
=$2\sqrt{3}+3\sqrt{3}-6\sqrt{3}-8\sqrt{3}$
=$-9\sqrt{3}$
b)$\sqrt{(2\sqrt{5}-7)^{2}}-\sqrt{45-20\sqrt{5}}$
=$\sqrt{(2\sqrt{5}-7)^{2}}-\sqrt{25-2·5·2\sqrt{5}+(2\sqrt{5})^{2}}$
=$\sqrt{(2\sqrt{5}-7)^{2}}-\sqrt{(5+2\sqrt{5})^{2}}$
=$|2\sqrt{5}-7|-|5+2\sqrt{5}|=7-2\sqrt{5}-5-2\sqrt{5}=2$
c)$\frac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\frac{2}{3}}+\frac{15}{\sqrt{6}-1}$
=$\frac{-2\sqrt{6}(5-\sqrt{6})}{5-\sqrt{6}}-\sqrt{\frac{9·2}{3}}+\frac{15}{\sqrt{6}-1}$
=$-2\sqrt{6}-\sqrt{6}+\frac{15}{\sqrt{6}-1}$
=$-3\sqrt{6}+\frac{15}{\sqrt{6}-1}$
=$-18+3\sqrt{6}+15=-3+3\sqrt{6}=3(-1+\sqrt{6})$
