#)Giải :

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)

\(=0\)

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)

\(=0\)

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\)    \(\left(1-\frac{1}{11}\right)\)\(\left(1-\frac{1}{13}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\)
 \(=1.\left(\frac{1}{3}-\frac{1}{2}\right).\left(\frac{1}{5}-\frac{1}{4}\right).\left(\frac{1}{7}-\frac{1}{6}\right)\)\(.\left(\frac{1}{11}-\frac{1}{10}\right).\frac{1}{13}\)
  \(=1.1.1.1.1.\frac{1}{13}\)
   \(=\frac{1}{13}\)
Cái này là chỉ do mình nghĩ thôi, ko biết đúng ko nữa
Có sai cko mình sorry nha!
Tk và kb hộ mình  với! thanks 

\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

Bạn tự viết lại đề bài nha

\(\frac{1}{2}\)x\(\frac{2}{3}\)x\(\frac{3}{4}\)x\(\frac{4}{5}\)x...x\(\frac{18}{19}\)x\(\frac{19}{20}\)

=\(\frac{1x2x3x4x...x18x19}{2x3x4x5x...x19x20}\)

=\(\frac{1}{20}\)

= \(\frac{1}{2}\). \(\frac{2}{3}\).\(\frac{3}{4}\).\(\frac{4}{5}\). ... . \(\frac{18}{19}\).\(\frac{19}{20}\)

= \(\frac{1}{2}\)

tk cho mk nha, mik đg âm điểm huhu

Đáp số là \(\frac{1}{20}\)

=>\(A=\frac{5-2}{5}.\frac{7-2}{7}.\frac{9-2}{9}....\frac{99-2}{99}\)

=>\(A=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}.....\frac{97}{99}=\frac{3.5.7.....97}{5.7.9.....99}=\frac{3.\left(5.7.9.....97\right)}{99.\left(5.7.9.....97\right)}=\frac{3}{99}=\frac{1}{3}\)

ĐKXĐ : \(x\ge0\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)

\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)

... 

\(A\le\frac{4.2010}{3}\) ma ban quan

Ta có : \(x^2+x+1\ge1\)vì \(x\ge0\)

Nên \(M=\frac{2020}{x^2+x+1}\le\frac{2020}{1}=2020\)

Vậy Max của M là 2020 khi x = 0

\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)

\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)

c)  C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )

     C = ( -1 ) + ( -1 ) + ... + ( -1 )

     C = ( -1 ) x ( 80 - 1 + 1 ) : 2

     C = ( -1 ) x 80 : 2

     C = ( -40 )

C=1-2+3-4+...+79-80

=(1-2)+(3-4)+...+(79-80)

=-1+(-1)+...+(-1) (có 80 số hạng bàng -1)

=-1*80

=-80