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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.
- \(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
- \(B=\left(\sqrt{28}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}=\left(2\sqrt{7}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}\)
\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)
- \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
- \(D=0,2.\sqrt{10^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{3}-\sqrt{5}\right)=4\sqrt{3}-2\sqrt{5}\)
a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}\)
\(=\left|\sqrt{7}+1\right|-\sqrt{7}\)
\(=\sqrt{7}+1-\sqrt{7}\)
=1
b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)
\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}\)
\(=\left|2+\sqrt{3}\right|-2\sqrt{3}\)
\(=2+\sqrt{3}-2\sqrt{3}\)
\(=2-\sqrt{3}\)
c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)
\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)
\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)
\(=\sqrt{13}-1+\sqrt{13}+1\)
\(=2\sqrt{13}\)
d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)
\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)
\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)
\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)
\(=\sqrt{21}-1-\sqrt{21}-1\)
=-2
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
a. \(\dfrac{3\sqrt{7}+7\sqrt{3}}{\sqrt{21}}=\dfrac{\sqrt{21}\left(\sqrt{3}+\sqrt{7}\right)}{\sqrt{21}}=\sqrt{7}+\sqrt{3}\)
b. \(\dfrac{2\sqrt{5}-4\sqrt{10}}{3\sqrt{10}}=\dfrac{\sqrt{10}\left(\sqrt{2}-4\right)}{3\sqrt{10}}=\dfrac{-4+\sqrt{2}}{3}\)
c. \(\dfrac{3-\sqrt{7}}{3+\sqrt{7}}-\dfrac{3+\sqrt{7}}{3-\sqrt{7}}=\dfrac{\left(3-\sqrt{7}\right)^2}{9-7}-\dfrac{\left(3+\sqrt{7}\right)^2}{9-7}=\dfrac{\left(3-\sqrt{7}-3-\sqrt{7}\right)\left(3-\sqrt{7}+3+\sqrt{7}\right)}{2}=\dfrac{-2\sqrt{7}.6}{2}=-6\sqrt{7}\)