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a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
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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\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
\(A=\sqrt{13+4\sqrt{10}}=\sqrt{13+2\sqrt{40}}=\sqrt{8+2.\sqrt{5}.\sqrt{8}+5}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=\sqrt{8}+\sqrt{5}\)
\(B=\sqrt{46-6\sqrt{5}}=\sqrt{46-2\sqrt{45}}=\sqrt{\left(\sqrt{45}-1\right)^2}=\sqrt{45}-1=3\sqrt{5}-1\)
\(C=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{5}+\sqrt{3}}{2}-\dfrac{\sqrt{7}+\sqrt{5}}{2}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{7}}{2}\)
\(C=\dfrac{-2\sqrt{3}-2\sqrt{2}+\sqrt{3}-\sqrt{7}}{2}=\dfrac{-\sqrt{3}-2\sqrt{2}-\sqrt{7}}{2}\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
\(13-2\sqrt{42}=7-2\sqrt{42}+6\\ =\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{7}-\sqrt{6}\right)^2\)
\(46+6\sqrt{5}=\left(5+2\cdot\sqrt{5}\cdot3+9\right)+32=\left(\sqrt{5}+3\right)^2+32\)(ko rút đc)
\(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\\ =\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{5-2\sqrt{5}+1}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\\ =4\left(3+\sqrt{5}\right)\)
\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Dễ dàng nhận ra
\(\sqrt{\sqrt{7}-\sqrt{3}}< \sqrt{\sqrt{7}+\sqrt{3}}\Rightarrow\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}< 0\)
Đặt \(x=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}< 0\)
\(\Rightarrow x^2=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}}{\sqrt{7}-2}\)
\(\Rightarrow x^2=\frac{2\sqrt{7}-2\sqrt{4}}{\sqrt{7}-2}=\frac{2\sqrt{7}-4}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
\(\Rightarrow x=-\sqrt{2}\) (do \(x< 0\))
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}\\ =\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{3+1-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\sqrt{3}-2}\\ =\sqrt{3+1+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-10\sqrt{3}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\\ =\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(D=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ \text{Ta có }:\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\\ =3+\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+3-\sqrt{5}\\ =6-2\sqrt{9-5}=6-2\sqrt{4}=6-4=2\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
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, \(\left(B-1\right)\left(B^2+B+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)