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a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}-1}\)
\(=\sqrt{4-\sqrt{5}}\)
c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=3-2=1\)
d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3+1}\)
\(=2\sqrt{3}\)
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
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\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}\\ =\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{3+1-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\sqrt{3}-2}\\ =\sqrt{3+1+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-10\sqrt{3}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\\ =\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(D=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ \text{Ta có }:\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\\ =3+\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+3-\sqrt{5}\\ =6-2\sqrt{9-5}=6-2\sqrt{4}=6-4=2\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
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, \(\left(B-1\right)\left(B^2+B+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
a,\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}=5+3\sqrt{2}\)
b, \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}+\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}\)
\(\Leftrightarrow\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}\)
\(\Leftrightarrow\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}-1+\sqrt{3}+1\)
\(\Leftrightarrow2\sqrt{3}\)
c, Ta có : \(\sqrt{13+4\sqrt{3}}=\sqrt{12+2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}+1\right)^2}=\sqrt{12}+1\)
=> \(\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)\(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)