c, C = 2/1.3 + 2/3.5 + 2/5.7 + .......+ 2/69.71+ 2/71.73
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a,0,36.350+1,2.20.3+9.4.4,5
=13.3.35+12.2.3+9.2.3.3
=3.(13.35+12.2+.9.2.3)
=3.(455+24+54)
=3.533
=1599
b,2015.2016-5/2015.2015+2010
=4062240-5+2010
=4064245
c,2/1.3+2/3.5+2/5.7+...+2/71.73
=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73
=1-1/73
=72/73
d,(1+1/2).(1+1/3)+...+(1+1/2018)
=3/2.4/3.5/4+...+2019/2018
=2019/2
e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn
=1/4-1/81
=77/324
f,F=3/2.3+3/3.4+...+3/99.100
=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d
=3.(1/2-1/100)
=3.49/100
=147/100
gG=5/1.4+5/4.7+...+5/61.64
3G=5.(3/1.4+3./4.7+...+3/61.64)
=5.(1-1/64)
=5.63/64
=315/64
ok nha bạn,mình giữ đúng lời hứa.
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
\(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{1}{55}\)
\(B=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}\)
\(B=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(B=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
ta có:
C = 1 - 1/3 + 1/3 - 1/5 +...+1/69 - 1/71 + 1/71 - 1/73
= 1 - 1/ 73
= 72/73
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{69.71}+\)\(\frac{2}{71.73}\)
\(C=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{69}-\frac{1}{71}+\frac{1}{71}-\frac{1}{73}\)
\(C=1-\frac{1}{73}\)
\(C=\frac{72}{73}\)