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f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
https://hoc24.vn/hoi-dap/question/407636.html
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}\)
= 9
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
D = (4\(\sqrt{10}\) - 4\(\sqrt{6}\) + 5\(\sqrt{6}\) - 3\(\sqrt{10}\) )\(\sqrt{4-\sqrt{15}}\)
D = (\(\sqrt{10}\) + \(\sqrt{6}\) )\(\sqrt{4-\sqrt{15}}\)
D = \(\sqrt{\left(4-\sqrt{15}\right)10}\) + \(\sqrt{\left(4-\sqrt{15}\right)6}\)
D = \(\sqrt{40-10\sqrt{15}}\) + \(\sqrt{24-6\sqrt{15}}\)
D = \(\sqrt{\left(\sqrt{15}\right)^2-2.5.\sqrt{5}+5^2}\) + \(\sqrt{\left(\sqrt{15}\right)^2-2.3.\sqrt{15}+3^2}\)
D = \(\sqrt{\left(\sqrt{15}-5\right)^2}\) + \(\sqrt{\left(\sqrt{15}-3\right)^2}\)
D = 5 - \(\sqrt{15}\) + \(\sqrt{15}\) - 3 = 2
b) \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}-\sqrt{2}.\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}-\sqrt{2}.\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}-\sqrt{10}+\sqrt{2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\sqrt{10}+\sqrt{2}=\dfrac{2\sqrt{5}}{\sqrt{2}}-\sqrt{10}+\sqrt{2}\)
\(=\sqrt{10}-\sqrt{10}+\sqrt{2}=\sqrt{2}\)
e) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(C=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
câu a ; f chưa nghỉ ra
\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)
\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)