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4 tháng 8 2018

Ta có  B = 7 − 5 2 3 + 19 − 6 2 = ( 1 − 2 ) 3 3 + ( 3 2 − 1 ) 2 = 1 − 2 + 3 2 − 1 = 2 2 .     

23 tháng 6 2019

a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)

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

b) Tương tự a) đ/s =5

5 tháng 10 2020

b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)

28 tháng 10 2020

a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)

\(=\frac{\sqrt{2\left(4-\sqrt{7}\right)}-\sqrt{2\left(4+\sqrt{7}\right)}+2}{\sqrt{2}}\)

\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\)

\(=\frac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+2}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|+2}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)+2}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-1-\sqrt{7}-1+2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)

b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)

\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)

\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)

\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)

26 tháng 7 2015

đăng ít thui má ạ 

25 tháng 10 2015

A=\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{5}+1-\sqrt{7+2\sqrt{10}}}\)=\(\frac{\sqrt{2}\left(\sqrt{3}+3+\sqrt{2}-\sqrt{5+2\sqrt{6}}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{5}+1-\sqrt{7+2\sqrt{10}}\right)}\)

A=\(\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{10+4\sqrt{6}}}{2+\sqrt{10}+\sqrt{2}-\sqrt{14+4\sqrt{10}}}=\frac{\sqrt{6}+3\sqrt{2}+2-\sqrt{6}-2}{2-\sqrt{10}+\sqrt{2}-\sqrt{10}-2}=\frac{3\sqrt{2}}{\sqrt{2}}=3\)

31 tháng 7 2019

mình nghĩ bài này sai đề, 

ĐÚng phải là\(\sqrt[3]{2+\sqrt{3}}\)

(   KHÔNG CHẮC NỮA   :D   )

1 tháng 8 2019

\(\text{sai đề chú ơi}\)

2 tháng 7 2018

a)                  \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\)\(\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{4+2\sqrt{3}}\)

                         \(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)

                          \(=\sqrt{5}-\sqrt{3}-\left(\sqrt{3}+1\right)=\sqrt{5}-1\)

\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)

b) tương tự câu a

c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}-\sqrt{6-2\sqrt{5+\sqrt{\left(\sqrt{12}-1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}-\sqrt{6-2\sqrt{5+\left(\sqrt{12}-1\right)}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}-\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}-\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\sqrt{6+2\left(\sqrt{3}-1\right)}-\sqrt{6-2\left(\sqrt{3}+1\right)}\)

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

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

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