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5 tháng 9 2020

a) Ta có: \(A=2\sqrt{2+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

        \(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\)

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

        \(\Leftrightarrow A=2\sqrt{2+\sqrt{5-\sqrt{12}+1}}\)

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

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

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

        \(\Leftrightarrow A=2\sqrt{\sqrt{3}+1}\)

        \(\Leftrightarrow A\approx3,30578\)

b) Ta có: \(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

        \(\Leftrightarrow B=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)

        \(\Leftrightarrow B=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)

        \(\Leftrightarrow B=\sqrt{2}.\left(4-2\right)\)

        \(\Leftrightarrow B=2\sqrt{2}\)

        \(\Leftrightarrow B\approx2,82843\)

5 tháng 8 2018

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

5 tháng 8 2018

a)  \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

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

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

b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)

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

\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)

c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)

14 tháng 6 2019

a) \(-\sqrt{3}\)      b) -10             c)  60               d)  -1             e) 1

 \(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}.6\sqrt{3}=36\)

\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)

\(=11-4\sqrt{6}\)

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

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

b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)

\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)

c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)

e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)

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

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

20 tháng 8 2019

a) \(\frac{2}{4-3\sqrt{2}}-\frac{2}{4+3\sqrt{2}}\)

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

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

\(=\frac{12\sqrt{2}}{-2}\)

\(=-6\sqrt{2}\)

b) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{4\sqrt{35}}{2}\)

\(=2\sqrt{35}\)

15 tháng 11 2019

a) 

= căn 5 -2

b)

\(\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\\ \)

\(\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)

\(2.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)

=2.(4-2)

=4.