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Mk nghĩ số 1000 ở mẫu bn nên sửa thành 2000 ms đúng
2014 × 2015 - 15/2000 + 2015 × 2013
= (2013 + 1) × 2015 - 15/2000 + 2015 × 2013
= 2013 × 2015 + (2015 - 15)/2000 + 2015 × 2013
= 2013 × 2015 + 2000/2000 + 2015 × 2013
= 1
=[2013*2014*+2014*2015+2015*2016]*[\(1\frac{1}{3}-1\frac{1}{3}\)]
=A*0
=0
nhớ k nha mấy friends/
\(\frac{2012+2013x2014}{2014x2015-2016}\)\(=\frac{2012+2013x2014}{2014x\left(2013+2\right)-2016}\)
\(=\frac{2012+2013x2014}{2014x2013+4028-2016}\)
\(=\frac{2012+2013x2014}{2014x2013+2012}\)
\(=1\)
\(\frac{2012+2013x2014}{2014x2015-2016}\)
= \(\frac{2012+2013x2014}{2014x\left(2013+2\right)-2016}\)
= \(\frac{2012+2013x2014}{2014x2013+4028-2016}\)
= \(\frac{2012+2013x2014}{2014x2013+2012}\)
= 1
\(\frac{2012+2013x2014}{2014x2015-2016}\)
= \(\frac{2012+4054182}{4058210-2016}\)
= \(\frac{4056194}{4056194}\)
= \(\frac{1}{1}\)
= \(1\)
Answer : \(1\)
\(\frac{2012+2013.2014}{2014.2015-2016}\)
\(=\frac{2012+2013.2014}{2014.\left(2013+2\right)-2016}\)
\(=\frac{2012+2013.2014}{2014.2013+2014.2-2016}\)
\(=\frac{2012}{4028-2016}\)
\(=\frac{2012}{2012}\)
\(=1\)
\(\frac{10}{2000}:\frac{10}{2000}.\frac{4}{5}+\frac{1}{5}+20\%+80\%\)
=\(1.\frac{4}{5}+\frac{1}{5}+\frac{1}{5}+\frac{4}{5}\)
=\(\frac{4}{5}.2+\frac{1}{5}.2\)
=\(\frac{8}{5}+\frac{2}{5}\)
=\(\frac{10}{5}=2\)
=
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
= \(\frac{13}{45}x\left(\frac{15}{26}-1\frac{1}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{15}{26}-\frac{27}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{27}{26}-\frac{27}{26}\right)=\frac{13}{45}x0=0\)
=13+2013x2014/2014x2013+2013-2000
=13+2014x2013/2014x2013+13
=1