Bạn cần nêu rõ yêu cầu đề bài là rút gọn.
Gọi biểu thức là $A$. Ta có:
$A\sqrt{2}=\sqrt{12-6\sqrt{3}}-\sqrt{12+6\sqrt{3}}+\sqrt{2}.\frac{4-2\sqrt{3}}{2-\sqrt{3}}$
$=\sqrt{9+3-2.3\sqrt{3}}-\sqrt{9+3+2.3\sqrt{3}}+\sqrt{2}.2$
$=\sqrt{(3-\sqrt{3})^2}-\sqrt{(3+\sqrt{3})^2}+\sqrt{2}.2$
$=3-\sqrt{3}-(3+\sqrt{3})+2\sqrt{2}$
$=-2\sqrt{3}+2\sqrt{2}$
$\Rightarrow A=-\sqrt{6}+2$
Bài tập:Rút gọn
1.\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
2. \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
3. \(8\sqrt{3}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
4.\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
5.(\(\sqrt{12}+\sqrt{75}+\sqrt{27}\)):\(\sqrt{15}\)
6.\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
2.
a,\(\sqrt{12}-\sqrt{27}+\sqrt{3}\)
b,\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448};\sqrt{3}.\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\)
c,\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}};\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
d,\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right).\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
\(B=2+3\sqrt{2}-2\sqrt{32}-\sqrt{6+4\sqrt{2}}\)
\(G=\sqrt{12-2\sqrt{35}}+4\sqrt{20}+\sqrt{28}\)
\(E=\left(\frac{2\sqrt{2}+3\sqrt{3}}{\sqrt{2}+\sqrt{3}}-\sqrt{6}\right):\left(\sqrt{2}-\sqrt{3}\right)-\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
1, \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{10}}\)
2, \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
3, \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
4, \(\sqrt{8\sqrt{3}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}}\)
5, \(\frac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-\sqrt{\frac{15}{2}}\)
ai giúp e với ạ
Rút gọn:
a. \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)
b.\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
Bài 3: Thực hiện phép tính
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\frac{10-2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
3)\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
5)\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
6)\(6\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
7)\(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
8)\(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
9)\(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(0.1\cdot\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
\(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
\(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
