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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{28-10\sqrt{3}}\\ =\sqrt{3-10\sqrt{3}+25}\\ =\sqrt{\left(\sqrt{3}-5\right)^2}\\ =\left|\sqrt{3}-5\right|\\ =5-\sqrt{3}\)
\(\sqrt{41+12\sqrt{5}}\\ =\sqrt{5+12\sqrt{5}+36}\\ =\sqrt{\left(\sqrt{5}+6\right)}\\ =\left|\sqrt{5}+6\right|\\ =\sqrt{5}+6\)
\(\sqrt{32-10\sqrt{7}}\\ =\sqrt{7-10\sqrt{7}+25}\\ =\sqrt{\left(\sqrt{7}-5\right)^2}\\ =\left|\sqrt{7}-5\right|\\ =5-\sqrt{7}\)
\(\sqrt{11-4\sqrt{7}}\\ =\sqrt{7-4\sqrt{7}+4}\\ =\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|\\ =\sqrt{7}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\sqrt{5}-2\sqrt{5}+9\sqrt{5}-30\sqrt{5}=-21\sqrt{5}\)
b: \(=2\sqrt{7}-6\sqrt{7}-\dfrac{3}{4}\sqrt{7}-8\sqrt{7}=-\dfrac{51}{4}\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)
câu tiếp tương tự câu thứ 2 nha
![](https://rs.olm.vn/images/avt/0.png?1311)
9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)
\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)
\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)
\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)
\(=-5\sqrt{3x}++27\)
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)
\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)
\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)
\(=13\sqrt{5a}+\sqrt{a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
\(\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\left(vì:\sqrt{2}>\sqrt{1}=1\right)\)
\(\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{\left(2\right)^2+2.2\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\left(2+\sqrt{3}>0\right)\)
\(\sqrt{28-10\sqrt{3}}=\sqrt{25-10\sqrt{3}+3}=\sqrt{5^2-2.5\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\left(vì:5=\sqrt{25}>\sqrt{3}\right)\)
Bạn giải thích cho mik chỗ \(\sqrt{\left(\sqrt{2}-1\right)^2}\)tại sao lại bằng Căn bậc 2 của 2 -1