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= \(\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+\frac{5.5}{4.6}+\frac{6.6}{5.7}\)
= \(\frac{2.3.4.5.6}{1.2.3.4.5}+\frac{2.3.4.5.6}{3.4.5.6.7}\)
= \(\frac{2}{1}+\frac{6}{7}\)
= 2\(\frac{6}{7}\)
Mình nghĩ zậy !!!!!!!!!!!!!!!!!!
Ta có:3/5+3/7-3/11=3.(1/5+1/7-1/11)
4/5+4/7-4/11=4.(1/5+1/7-1/11)
=>M=[3.(1/5+1/7-1/11)]/[4.(1/5+1/7-1/11)]=3/4
M = \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{3}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{2009}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
Ta có: \(A = 2{\sin ^2}\alpha + 5{\cos ^2}\alpha = 2({\sin ^2}\alpha + {\cos ^2}\alpha ) + 3{\cos ^2}\alpha \)
Mà \({\cos ^2}\alpha + {\sin ^2}\alpha = 1;\cos \alpha = - \frac{{\sqrt 2 }}{2}.\)
\( \Rightarrow A = 2 + 3.{\left( { - \frac{{\sqrt 2 }}{2}} \right)^2} = 2 + 3.\frac{1}{2} = \frac{7}{2}.\)
- Xét \(sin\frac{x}{5}=0\Rightarrow C=...\)
- Với \(sin\frac{x}{5}\ne0\)
\(C.sin\frac{x}{5}=sin\frac{x}{5}.cos\frac{x}{5}.cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{2}sin\frac{2x}{5}cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{4}sin\frac{4x}{5}cos\frac{4x}{5}cos\frac{8x}{5}=\frac{1}{8}sin\frac{8x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{16}sin\frac{16x}{5}\Rightarrow C=\frac{sin\frac{16x}{5}}{16.sin\frac{x}{5}}\)
\(D=sin\frac{x}{7}+sin\frac{5x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}cos\frac{2x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}\left(cos\frac{2x}{7}+1\right)=4cos^2\frac{x}{7}.sin\frac{3x}{7}\)
\(A=cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7}=cos\frac{\pi}{7}cos\frac{4\pi}{7}cos\frac{2\pi}{7}\)
\(\Rightarrow A.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{2}sin\frac{2\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{8}sin\frac{8\pi}{7}=\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=-\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=-\frac{1}{8}\)
\(B=sin6.cos48.cos24.cos12\)
\(B.cos6=sin6.cos6.cos12.cos24.cos48\)
\(=\frac{1}{2}sin12.cos12.cos24.cos48=\frac{1}{4}sin24.cos24.cos48\)
\(=\frac{1}{8}sin48.cos48=\frac{1}{16}sin96\)
\(=\frac{1}{16}sin\left(90+6\right)=\frac{1}{16}cos6\Rightarrow B=\frac{1}{16}\)
\(cosx+cos\left(x+\frac{\pi}{5}\right)+cos\left(x+\frac{9\pi}{5}\right)+cos\left(x+\frac{2\pi}{5}\right)+cos\left(x+\frac{8\pi}{5}\right)+...+cos\left(x+\frac{5\pi}{5}\right)\)
\(=cosx-2cosx.cos\frac{4\pi}{5}-2cosx.cos\frac{3\pi}{5}-2cosx.cos\frac{2\pi}{5}-2cosx.cos\frac{\pi}{5}-cosx\)
\(=-2cosx\left(cos\frac{\pi}{5}+cos\frac{4\pi}{5}+cos\frac{2\pi}{5}+cos\frac{3\pi}{5}\right)\)
\(=-2cosx\left(2cos\frac{\pi}{2}.cos\frac{3\pi}{10}+2cos\frac{\pi}{2}cos\frac{\pi}{10}\right)\)
\(=0\) (do \(cos\frac{\pi}{2}=0\))
Ta có:
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{91.93}+\frac{5}{93.95}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}+\frac{1}{93.95}\right)=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}+\frac{2}{93.95}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}+\frac{1}{93}-\frac{1}{95}\right)=\frac{5}{2}\left(1-\frac{1}{95}\right)=\frac{5}{2}.\frac{94}{95}=\frac{47}{19}\)
Vậy \(A=\frac{47}{19}\)
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{93.95}\)
\(A=5\cdot\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{95}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{95}\right)=\frac{5}{2}\cdot\frac{94}{95}=\frac{47}{19}\)