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a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
b,c tương tự
\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)ta có:
\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3B\sqrt[3]{25-52}\)
\(=10-9B\)
Giải PT: \(B^3+9B-10=0\Leftrightarrow B^3-1+9B-9=0\)\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+1\right)+9\left(B-1\right)=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+10\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}B-1=0\\B^2+2B+1+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+1\right)^2=-9\left(L\right)\end{cases}}}\)
Vậy \(B=1\)
À chết mình làm nhầm, phải là \(\left(B-1\right)\left(B^2+B+1\right)\) nha, \(\left(B-1\right)\left(B^2+B+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)
\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2}+4\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{5^2+2.5.3\sqrt{2}+\left(3+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+3+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+6\right)}=\sqrt{11}\)
\(=5+6=11\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
mik chỉnh lại đề
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
\(b,=\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) \(=\sqrt{3+30\sqrt{2+\sqrt{8+2\sqrt{8}+1}}}\)
\(=\sqrt{3+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)\(=\sqrt{3+30\sqrt{3+\sqrt{8}}}=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{3+30\sqrt{2}+30}=\sqrt{33+30\sqrt{2}}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1
b) Ta có: \(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{33+30\sqrt{2}}\)