Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)

\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)

\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)

B=2012.A

=>A/B=1/2012

a/b= 1/2012 nha bạn 

tích

kobiet

a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)

Ta có: \({\sin ^2}a + {\cos ^2}a  = 1\)

 \(\Leftrightarrow \frac{1}{9} + {\cos ^2}a  = 1\)

\(\Leftrightarrow {\cos ^2}a =  1 - \frac{1}{9}= \frac{8}{9}\)

\(\Leftrightarrow \cos a  =\pm\sqrt { \frac{8}{9}}  =  \pm \frac{{2\sqrt 2 }}{3}\)

Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)

Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} =  - \frac{{\sqrt 2 }}{4}\)

Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\)

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)

\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} =  - \frac{{4\sqrt 2 }}{7}\)

b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)

\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)

Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 =  - \frac{3}{4}\)

Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)

\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)

\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)

\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 =  - \frac{{\sqrt 7 }}{4}\)

\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)

mình cần gấp

Ta có: \(A=1\times2010+2\times2009+\cdots+2010\times1\)

\(=2\times\left(1\times2010+2\times2009+\cdots+1005\times1006\right)\)

\(=2\times\left\lbrack1\times\left(2011-1\right)+2\times\left(2011-2\right)+\cdots+1005\times\left(2011-1005\right)\right\rbrack\)

\(=2\times\left\lbrack2011\left(1+2+\cdots+1005\right)-\left(1\times1+2\times2+\ldots+1005\times1005\right)\right\rbrack\)

\(=2\times\left\lbrack2011\times1005\times\frac{1006}{2}-\frac{1005\times\left(1005+1\right)\times\left(2\times1005+1\right)}{6}\right\rbrack\)

\(=2\left\lbrack2011\times1005\times\frac{1006}{2}-\frac{1005\times1006\times2011}{6}\right\rbrack\)

\(=2011\times1005\times1006-\frac{1005\times1006\times2011}{3}\)

\(=2011\times1005\times1006-335\times1006\times2011=2011\times1006\times335\left(3-1\right)=2\times335\times1006\times2011\)

=335x2011x2012

Ta có: B=1+(1+2)+(1+2+3)+...+(1+2+3+...+2010)

\(=\frac{1\times2}{2}+\frac{2\times3}{2}+\cdots+\frac{2010\times2011}{2}\)

\(=\frac12\times\left(1\times2+2\times3+\cdots+2010\times2011\right)\)
\(=\frac12\times\left\lbrack1\times\left(1+1\right)+\right.2\times\left(2+1)+\cdots+2010\times\left(2010+1\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+2010\times2010\right)+\left(1+2+\cdots+2010\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2010\times\left(2010+1\right)\left(2\times2010+1\right)}{6}+\frac{2010\times2011}{2}\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2010\times2011\times4021}{6}+\frac{2010\times2011}{2}\right\rbrack=\frac12\times\frac{2010\times2011\times4021+3\times2010\times2011}{6}\)

\(=\frac12\times\frac16\times2010\times2011\times\left(4021+3\right)=\frac12\times\frac16\times2010\times2011\times4024\)

\(=2010\times2011\times\frac{2012}{6}=2011\times2012\times335\)

=A

=>A:B=1

Theo đề,  ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=162\\\dfrac{1}{2}a-\dfrac{1}{3}b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=66\\b=96\end{matrix}\right.\)