1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)

\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)

\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)

2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)

\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)

\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)

3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)

\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)

4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)

\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)

\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)

6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)

\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)

=-3-9

=-12

5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)

7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)

\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)

\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)

8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)

\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)

9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)

\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)

\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)

10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)

\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)

\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)

11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)

3.728493106

mk trước

2.

A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)

\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}\)

\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)

\(=\sqrt{3}-1-\sqrt{3}-1\)

\(=-2\)

B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)

\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{5-2\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)

\(=\sqrt{5-2\sqrt{2}-2}\)

\(=\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}\)

\(=\sqrt{2}-1\)

\(\sqrt{28-6\sqrt{3}}\) ms đúng đề chứ bạn

\(\sqrt{28-16\sqrt{3}}+\sqrt{13-4\sqrt{3}}\)

\(=\sqrt{\left(4-2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-1\right)^2}\)

\(=\left|4-2\sqrt{3}\right|+\left|2\sqrt{3}-1\right|\)

\(=4-2\sqrt{3}+2\sqrt{3}-1=3\)

\(\sqrt{48}=4\sqrt{3}\) =>7+\(\sqrt{48}=7+4\sqrt{3}=\)(\(2+\sqrt{3}\))²

\(\sqrt{28-16\sqrt{3}}=2\sqrt{7-4\sqrt{3}}\)=2(2-\(\sqrt{3}\))=4-2\(\sqrt{3}\)=(\(\sqrt{3}-1\))²

viết lại biểu thức ta được

(\(\sqrt{\sqrt{7}+4\sqrt{3}}-\left(\sqrt{3}-1\right)\))(2+\(\sqrt{3}\))

Xem lại đề bài?