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Ta có: \(B=\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+3\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+3\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+3\cdot\left(\sqrt{2}+1\right)}\)
\(=\sqrt{13+3\sqrt{2}+3}\)
\(=\sqrt{16+3\sqrt{2}}\)
Ta có: \(C=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\left(\sqrt{3}+1\right)^2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{\sqrt{3}-1}=1\)
Ta có: \(D=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-2}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{1}}\)
\(=\sqrt{6+2\sqrt{2}}\)
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)
\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
\(A=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{8-\sqrt{48}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6-2\sqrt{6.2}+2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}=1\)
Đề thiếu nha:
\(E=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)(vì \(\sqrt{3}>1\))
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Lời giải:
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}\)
\(=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}\)
\(=\sqrt{3}+1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}\)
\(=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1\)
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\\ =\sqrt{6-2\sqrt{3}-2}\\ =\sqrt{4-2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\\ =\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\\ =\frac{2}{\sqrt{3}-1}+\sqrt{3}\\ =\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\frac{5-\sqrt{3}}{\sqrt{3}-1}\\ =\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\\ -\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4\)
(bạn nhớ ktr đã nha)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2-\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{2}.\sqrt{4-2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=1\)