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a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)
\(=\left(3-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2\right)\)
\(=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)=9-3=6\)
\(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}=3+\frac{1-\sqrt{3}+2}{1-\sqrt{3}}=3+1+\frac{2}{1-\sqrt{3}}=4+\frac{2}{1-\sqrt{3}}\)
\(=4+\frac{3-1}{1-\sqrt{3}}=4+\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{1-\sqrt{3}}=4-\sqrt{3}-1=-\sqrt{3}-3\)
\(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2=\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2=\left(\sqrt{3}+3\right)\)
Khi đó \(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)=-\left(\sqrt{3}+3\right)^2=-12-6\sqrt{3}\)