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c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)
=> đpcm.
\(A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}+1}}=\dfrac{1-\sqrt{x-1}}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)
\(A=\dfrac{1-\sqrt{x-1}}{\left|\sqrt{x-1}-1\right|}\) \(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}=-1\\A=\dfrac{1-\sqrt{x-1}}{-\sqrt{x-1}+1}=1\end{matrix}\right.\)
\(B=\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(B^2=\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)^2\)
\(B^2=8+2\sqrt{10+2\sqrt{5}}-2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}+8-2\sqrt{10+2\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(B^2=16-2\sqrt{8^2-4\left(10+2\sqrt{5}\right)}\)
\(B^2=16-2\sqrt{24-8\sqrt{5}}\)
\(B^2=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(B^2=16-4\sqrt{5}+4=20-4\sqrt{5}\)
\(B=\sqrt{20-4\sqrt{5}}\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
\(a.\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
\(b.\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=\sqrt{2}+5-\left(\sqrt{2}-5\right)=\sqrt{2}+5-\sqrt{2}+5=10\)
- Tự làm mấy câu còn lại cho quen đi b
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
\(a.B=\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
\(b.A=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\dfrac{1}{\sqrt{2}+1}-\sqrt{2}=\dfrac{-1-\sqrt{2}}{\sqrt{2}+1}=-1\)
\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\) \(=\frac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)
=\(2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
a) Ta có: \(\sqrt{3+2\sqrt{2}-\sqrt{3-2\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}-\left|\sqrt{2}-1\right|}\)
\(=\sqrt{3+2\sqrt{2}-\left(\sqrt{2}-1\right)}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{2}+1}\)
\(=\sqrt{4+\sqrt{2}}\)
b) Ta có: \(\sqrt{7-4\sqrt{3}+\sqrt{12+6\sqrt{3}}}\)
\(=\sqrt{7-4\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{3}\cdot3}}\)
\(=\sqrt{7-4\sqrt{3}+\sqrt{\left(3+\sqrt{3}\right)^2}}\)
\(=\sqrt{7-4\sqrt{3}+\left|3+\sqrt{3}\right|}\)
\(=\sqrt{7-4\sqrt{3}+3+\sqrt{3}}\)
\(=\sqrt{10-3\sqrt{3}}\)
c) Ta có: \(\sqrt{5-2\sqrt{6}}+\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{5}\right|\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{5}\)
\(=\sqrt{3}+\sqrt{5}\)
d) Ta có: \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\sqrt{6-2\cdot\sqrt{6}\cdot\sqrt{2}+2}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\left|\sqrt{6}-\sqrt{2}\right|}{\sqrt{3}-1}-2\sqrt{2}\)
\(=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-2\sqrt{2}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-2\sqrt{2}\)
\(=2-2\sqrt{2}\)
mình chép ra giấy
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}+8-2\sqrt{10+2\sqrt{5}}\)
\(=16+2\sqrt{24-2.2.2\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\left(2\sqrt{5}-2\right)=16+4\sqrt{5}-4=12+4\sqrt{5}\)
Vậy \(A=\sqrt{12+4\sqrt{5}}=\sqrt{4.3+4\sqrt{5}}=2\sqrt{3+\sqrt{5}}\)