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2:

A=-(1/4-1/5+1/5-1/6+...+1/9-1/10)

=-(1/4-1/10)

=-1/4+1/10

=-5/20+2/20=-3/20

19 tháng 4 2019

nhầm:mux3=mũ3

là số nguyên tố 

 

27 tháng 7 2016

\(\left(\frac{1999}{2011}-\frac{2011}{199}\right)-\left(-\frac{12}{1999}-\frac{12}{2001}\right)\)

=\(\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)

=\(\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(-\frac{2011}{1999}+\frac{12}{1999}\right)\)

=\(\frac{2011}{2011}+\frac{1999}{1999}\)

=1+1

=2

27 tháng 7 2016

\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)

=\(\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)

=\(\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{-2011}{1999}+\frac{12}{1999}\right)\)

=1-1

=-2

4 tháng 5 2017

A.-9/10*5/14*1/10*(-9/2)+1/7*(-9/10)

=-9/10*(5/14+1/7)*1/10*(-9/2)

=-9/10*1/2*1/10*(-9/2)

=-9/20*1/10*(-9/10)

=-9/200*-9/10

=81/2000

B. (1999/2011-2011/1999)-(-12/1999-12/2011)

    =1999/2011-2011/1999+12/1999+12/2011

    =(1999/2011+12/2011)-(2011/1999-12/1999)

    = 1-1

    =0

DD
2 tháng 6 2021

a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)

b) Bạn làm tương tự. 

7 tháng 2 2016

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+....+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)

7 tháng 2 2016

b.=3/2.4/3....2012/2011

=3.4....2012/2.3....2011=2012/2=1006

24 tháng 10 2020

\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)

Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó

\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)

\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)

24 tháng 10 2020

Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)

\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)

\(=\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)

\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

Ta có: \(F\left(x\right)+F\left(1-x\right)\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)

\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)

...

\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)

Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)

\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)

\(=1+1+...+F\left(\frac{1}{2}\right)\)

\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)

\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)

\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)

\(=1005+\frac{1}{2}=\frac{2011}{2}\)

20 tháng 12 2015

phantuananh mấy tháng nữa chắc mk cũng chả cần nữa rồi