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A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
Đặt \(\left\{{}\begin{matrix}x-2008=n\\2x+2009=h\\3x-2011=t\end{matrix}\right.\Rightarrow n+h+t=6x-2010\)
\(\Rightarrow pt\Leftrightarrow\dfrac{1}{n}+\dfrac{1}{h}=\dfrac{1}{n+h+t}-\dfrac{1}{t}\)
\(\Leftrightarrow\dfrac{n+h}{hn}=\dfrac{-\left(n+h\right)}{t\left(n+h+t\right)}\)
\(\Leftrightarrow\left(n+h\right)\left(\dfrac{1}{hn}+\dfrac{1}{t\left(n+h+t\right)}\right)=0\)
\(\Leftrightarrow\left(n+h\right)\dfrac{t\left(n+h+t\right)+hn}{hnt\left(n+h+t\right)}=0\)
\(\Leftrightarrow\dfrac{\left(n+h\right)\left(n+t\right)\left(t+h\right)}{hnt\left(n+h+t\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}n=-h\\n=-t\\t=-h\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-2008=-\left(2x+2009\right)\\x-2008=-\left(3x-2011\right)\\3x-2011=-\left(2x+2009\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{4019}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)
Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
Suy luận.
Tử số của P lớn hơn hoặc bằng 2, còn mẫu số là sin 2 a . cos 2 a = 2/3. 1/3 = 2/9 < 1/4, nên P ≤ 8. Do đó các phương án A, B, D bị loại. Đáp án là C.
a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)
b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)
c/\(P=sin\left(30+60\right)=sin90=1\)
d/
\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)
\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)
e/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)
\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)
\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)
\(=\dfrac{123}{456}\left(1-2\right)\)
\(=-\dfrac{123}{456}\)
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 = \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{3}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)
C=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009-2010-2011
=-1-2011
=-2012