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vì 1/1*2=1-1/2
1/2*3=1/2-1/3
.....................
1/2014*2015=1/2014-1/2015
=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015
=1-1/2015
=2014/2115
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
\(t=\frac{2009.2010+2000}{2011.2010-2020}\)
\(t=\frac{2009.2010+2000}{\left(2009+2\right).2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2.2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2000}=1\)
\(q=\frac{\text{2014.2015+2010}}{2016.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\left(2014+2\right).2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{2014.2015+2.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\text{2014.2015+2010}}=1\)
- 225x17-225x16-225=225x[17-16-1]
=225x0
=0
- 28x76+24x28-2x48x18=28x[76+28] -2x48x18
= 28x104-1728
= 291-1728
=1184
Ta có: \(\frac{2015.2016-2}{2014.2015+4028}=\frac{2015.2016-2}{2014.2015+2.2014}=\frac{\left(2015+2\right).\left(2016-2\right)}{2014.\left(2015+2\right)}=\frac{2017.2014}{2017.2014}=1\)