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24 tháng 4 2019

4/3.5+4/5.7+...+4/98.100(sửa vì  đề sai)

=2(1/3-1/5+1/5-1/7+...+1/98-1/100)

=2(1/3-1/100)

=2.(97/300)

=97/150

24 tháng 4 2019

\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.100}\)

\(=2.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.100}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{100}\right)\)

\(=2.\frac{97}{300}\)

\(=\frac{97}{150}\)

23 tháng 4 2019

Đặt \(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{101}{303}-\frac{3}{303}\)

\(\Rightarrow\frac{1}{2}A=\frac{98}{303}\)

\(\Rightarrow A=\frac{98}{303}\div\frac{1}{2}\)

\(\Rightarrow A=\frac{199}{303}\)

23 tháng 4 2019

\(x+\frac{3}{22}=\frac{27}{121}.\frac{11}{9}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{27.11}{121.9}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{3.1}{11.1}\)

\(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)

\(\Leftrightarrow x=\frac{3}{11}-\frac{3}{22}\)

\(\Leftrightarrow x=\frac{6}{22}-\frac{3}{22}\)

\(\Leftrightarrow x=\frac{3}{22}\)

1 tháng 4 2017

\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+...+\frac{4}{97\cdot99}\)

\(=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+...+\frac{2\cdot2}{97\cdot99}\)

\(=\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+\frac{2}{7}-\frac{2}{7}+\frac{2}{9}-...+\frac{2}{97}-\frac{2}{99}\)

\(=\frac{2}{3}-\frac{2}{99}\)

\(=\frac{64}{99}\)

19 tháng 9 2021

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.95}\)

\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)\)

\(=2.\dfrac{94}{291}=\dfrac{188}{291}\)

19 tháng 9 2021

\(=2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)=2\cdot\dfrac{94}{291}=\dfrac{188}{291}\)

8 tháng 8 2023

 

a/

\(A=3^2+3^2.2^2+3^2.3^2+3^2.4^2+...+3^2.30^2=\)

\(=3^2\left(1^2+2^2+3^2+...+30^2\right)\)

Đăt biểu thức trong dấu ngoặc là B

\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+30\left(31-1\right)=\)

\(=1.2+2.3+3.4+30.31-\left(1+2+3+...+30\right)=\)

\(C=1+2+3+...+30=\dfrac{30\left(1+30\right)}{2}=465\)

\(D=1.2+2.3+3.4+...+30.31\)

\(3D=1.2.3+2.3.3+3.4.3+...+30.31.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+30.31.\left(32-29\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-29.30.31+30.31.32=\)

\(=30.31.32\Rightarrow D=\dfrac{30.31.32}{3}=10.31.32\)

\(\Rightarrow A=3^2\left(C-D\right)=3^2\left(10.31.32-465\right)\)

b/

Đặt biểu thức là A

\(\dfrac{A}{2}=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{39-37}{37.39}=\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}-\dfrac{1}{39}=\)

\(=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}\Rightarrow A=\dfrac{2.12}{39}=\dfrac{24}{39}=\dfrac{8}{13}\)

8 tháng 8 2023

\(A=3.3+6.6+9.9+...+90.90\)

\(A=3^2\left(1+2^2+3^2+...+10^2\right)\)

\(A=9.\dfrac{10.\left(10+1\right)\left(2.10+1\right)}{6}\)

\(A=3.\dfrac{10.11.21}{2}\)

\(A=3465\)

26 tháng 4 2017

\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)

= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)

=1+       0          +        0        +        0         +...+        0          +         \(\frac{1}{2013}\)

=1+\(\frac{1}{2013}\)

=\(\frac{2014}{2013}\)

k dùm nha

26 tháng 4 2017

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)

\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=2\cdot\left(1-\frac{1}{2013}\right)\)

\(=2\cdot\frac{2012}{2013}\)

\(=\frac{4024}{2013}\)

11 tháng 4 2022

T=4/1 . 4/3 + 4/3 . 4/5 + ... + 4/99 . 4/100

T=4/1 - 4/3 + 4/3 - 4/5 + ... + 4/99 - 4/100

T=4/1 - 4/100

T=99/25

11 tháng 4 2022

hỏi văn chấm?

giải cả cách lm nx xem nào a:> ?

7 tháng 5 2015

\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2013.2015}=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=2.\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)

\(=2.\frac{2014}{2015}\)

\(=\frac{4028}{2015}\)