a) Ta có:
\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=\dfrac{1}{2}\sqrt{16. 3}-2\sqrt{25. 3}-\dfrac{\sqrt{3.11}}{\sqrt{11}}+5\sqrt{\dfrac{1.3+1}{3}}\)
\(=\dfrac{1}{2}\sqrt{4^2. 3}-2\sqrt{5^2. 3}-\dfrac{\sqrt 3.\sqrt{11}}{\sqrt{11}}+5\sqrt{\dfrac{4}{3}}\)
\(=\dfrac{1}{2}.4\sqrt{ 3}-2.5\sqrt{3}-\sqrt{3}+5\dfrac{\sqrt 4}{\sqrt 3}\)
\(=\dfrac{4}{2}\sqrt{ 3}-10\sqrt{3}-\sqrt{3}+5\dfrac{\sqrt{4}.\sqrt 3}{\sqrt{3}.\sqrt {3}}\)
\(=2\sqrt{ 3}-10\sqrt{3}-\sqrt{3}+5\dfrac{2\sqrt{3}}{3}\)
\(=2\sqrt{ 3}-10\sqrt{3}-\sqrt{3}+10\dfrac{\sqrt{3}}{3}\)
\(= \left( {2 - 10 - 1 + \dfrac{10}{3} }\right)\sqrt 3 \)
\(=-\dfrac{17}{3}\sqrt 3\).
b) Ta có:
\(\sqrt{150}+\sqrt{1,6}. \sqrt{60}+4,5. \sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
\(=\sqrt{25. 6}+\sqrt{1,6. 60}+4,5.\sqrt{\dfrac{2.3+2}{3}}-\sqrt{6}\)
\(=\sqrt{5^2. 6}+\sqrt{1,6. (6.10)}+4,5\sqrt{\dfrac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{ 6}+\sqrt{(1,6. 10).6}+4,5\dfrac{\sqrt 8}{\sqrt 3}-\sqrt{6}\)
\(=5\sqrt{ 6}+\sqrt{16.6}+4,5\dfrac{\sqrt 8 . \sqrt 3}{ 3}-\sqrt{6}\)
\(=5\sqrt{ 6}+\sqrt{4^2.6}+4,5\dfrac{\sqrt {8 .3}}{ 3}-\sqrt{6}\)
\(= 5\sqrt{6}+4\sqrt{ 6}+4,5. \dfrac{\sqrt{4.2. 3}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+4,5. \dfrac{\sqrt{2^2.6}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+4,5. 2\dfrac{\sqrt{6}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+9\dfrac{\sqrt{6}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}\)
\(=(5+4+3-1)\sqrt{6}=11\sqrt{6}.\)
Cách 2: Ta biến đổi từng hạng tử rồi thay vào biểu thức ban đầu:
+ \(\sqrt{150}=\sqrt{25.6}=5\sqrt 6\)
+ \(\sqrt{1,6.60}=\sqrt{1,6.(10.6)}=\sqrt{(1,6.10).6}=\sqrt{16.6}\)
\(=4\sqrt 6\)
+ \(4,5.\sqrt{2\dfrac{2}{3}}=4,5.\sqrt{\dfrac{2.3+2}{3}}=4,5.\sqrt{\dfrac{8}{3}}=4,5\dfrac{8.3}{3}\)
\(=4,5.\dfrac{\sqrt{4.2.3}}{3}=4,5.\dfrac{2.\sqrt 6}{3}=9.\dfrac{\sqrt 6}{3}=3\sqrt 6.\)
Do đó:
\(\sqrt{150}+\sqrt{1,6}. \sqrt{60}+4,5. \sqrt{2\dfrac{2}{3}}-\sqrt{6}\)
\(=5\sqrt 6+4\sqrt 6+3\sqrt 6 - \sqrt 6\)
\(=(5+4+3-1)\sqrt 6=11\sqrt{6}\)
c) Ta có:
\(=(\sqrt{28}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{84}\)
\(=(\sqrt{4.7}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{4.21}\)
\(=(\sqrt{2^2.7}-2\sqrt{3}+\sqrt{7})\sqrt{7}+\sqrt{2^2.21}\)
\(=(2\sqrt{7}-2\sqrt{3}+\sqrt{7})\sqrt{7}+2\sqrt{21}\)
\(= 2\sqrt{7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+2\sqrt{21}\)
\(=2.(\sqrt{7})^2-2\sqrt{3.7}+(\sqrt{7})^2+2\sqrt{21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7=21\).
d) Ta có:
\((\sqrt{6}+\sqrt{5})^{2}-\sqrt{120}\)
\(=(\sqrt 6)^2+2.\sqrt 6 .\sqrt 5+(\sqrt 5)^2-\sqrt{4.30}\)
\(=6+2\sqrt{6.5}+5-2\sqrt{30}\)
\(=6+2\sqrt{30}+5-2\sqrt{30}=6+5=11.\)
