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AH
Akai Haruma
Giáo viên
20 tháng 7 2022

Lời giải:

a. $0,5.\frac{1}{2}+\frac{3}{4}=\frac{1}{4}+\frac{3}{4}=1$
b.

$\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{27.29}$

$=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{29-27}{27.29}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{29}$

$=1-\frac{1}{29}=\frac{28}{29}$

26 tháng 8 2019

Tính :

a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

26 tháng 8 2019

thanks

27 tháng 5 2022

\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{29\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ =\dfrac{1}{1}-\dfrac{1}{31}\\ =\dfrac{30}{31}\)

mà \(\dfrac{30}{31}>\dfrac{2014}{2015}\Rightarrow S>P\)

 

27 tháng 5 2022

So sánh vs j nhỉ .-.?

`S=1-1/3+1/3-1/5+...+1/29-1/31`

`S=1-1/31=30/31`

\(S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}=1-\dfrac{1}{31}=\dfrac{30}{31}\)

P=2014/2015=1-1/2015

mà 1/31>1/2015

nên S<P

18 tháng 5 2022

thank you cj

\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)

\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)

\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)

\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)

\(=-\frac{28}{15}\)

6 tháng 3 2016

tks bạn nha ^_^ Miu Ti

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

8 tháng 5 2017

Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{50}{101}\)

b) \(\frac{2^{10}+3^{31}+2^{40}+3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}=\frac{2^{10}+2^{40}}{2^{11}+2^{41}}\)

\(\frac{2^{10}+2^{40}}{2^{11}+2^{41}}=\frac{1}{2}\)

8 tháng 5 2017

=1/2x(1/1.3+1/3.5+...+1/99.101)

=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)

=1/2.(1-1/101)

=1/2.100/101

=50/101

chúc bạn học tốt

10 tháng 3 2016

\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

10 tháng 3 2016

2*(1/1*3+1/3*5+.......+1/99*100)

=2*(2/1*3+2/3*5+.....+2/99*100)*1/2

=1/3-1/5+1/5-1/7+....+1/99-1/100

=1/3-1/100

=100/300-3/300

=97/300

18 tháng 5 2018

1)           P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15

              P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)

              P=1/3-1/15= 4/15

2) a/ 0,2:1+3/5+80%

= 2/10:8/5+8/10

= 2/10.5/8+8/10

= 1/8 + 4/5 = 5/40 + 32/40 = 37/40

    b/ 0,5:5/4-2+1/5

= 5/10:5/4-11/5

= 5/10.4/5-11/5

=2/5-11/5 = -9/5

14 tháng 4 2016

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