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Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}=2\sqrt{3+\sqrt{3}-1}\)
\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
Suy ra VT = VP = 1
\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2.\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}=0\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
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) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)
\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
Đặt \(A=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
\(A=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{4}.\sqrt{12}}}}\)
\(A=\sqrt{6+2\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}\)
\(A=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)
\(A=\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)
\(A=\sqrt{6+2\sqrt{4-\sqrt{12}}}\)
\(A=\sqrt{6+2\sqrt{4-\sqrt{4}.\sqrt{3}}}\)
\(A=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)
\(A=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(A=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(A=\sqrt{6+2\sqrt{3}-2}\)
\(A=\sqrt{4+2\sqrt{3}}\)
\(A=\sqrt{3+2\sqrt{3}+1}\)
\(A=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(A=1+\sqrt{3}\) (đpcm)
Vậy \(A=1+\sqrt{3}\)
Ta có :
A= \(\dfrac{2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Đặt B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Ta có B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}}\)
\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12}-1}}\\ =2\sqrt{3+\sqrt{4-\sqrt{12}}}\\ =2\cdot\sqrt{3+\sqrt{3-2\cdot\sqrt{3}+1}}\\ =2\cdot\sqrt{3+\sqrt{3}-1}\\ =2\cdot\sqrt{2+\sqrt{3}}\)
Thay B vào A, ta cũng có:
A=\(\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\ =\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2\cdot\left(\sqrt{3}+1\right)}}\\ =\dfrac{\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}=1\)
Vậy A thuộc Z
Sửa đề: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
Ta có: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|-\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}-\sqrt{2}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{2}-\sqrt{3}\)
\(=-2\sqrt{2}\)(đpcm)
Biến đổi vế trái ta có:
\(\sqrt{3+\sqrt{5}-\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{5}-\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{3+\sqrt{5}-2\sqrt{3}-1}=\sqrt{2+\sqrt{5}-2\sqrt{3}}\)
Đề sai