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26 tháng 4 2017

Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

16 tháng 4 2016

có nhầm đề ko? cái chỗ 3/92.98

16 tháng 4 2016

\(A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot98}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}+...+\frac{1}{92}-\frac{1}{98}\)

\(A=\frac{1}{2}-\frac{1}{98}\)

\(A=\frac{47}{98}\)

TÍCH CHO MK NHA

31 tháng 3 2019

1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

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

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

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

d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)

\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)

\(D=\frac{1}{2}-\frac{1}{95}\)

\(D=\frac{95-2}{190}=\frac{93}{190}\)

Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)

6 tháng 9 2020

a) 2/2.5 + 2/5.8 + 2/8.11 + ... + 2/x(x+3) = 7/23

3/2.5 + 3/5.8 + 3/8.11 + ... + 3/x(x+3) = 21/46

1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x+1 = 21/46

1/2 - 1/x+1 = 21/46

=> 1/x+1 = 1/23

=> x + 1 = 23

=> x = 22

Vậy x = 22.

b) 3/4 . x - 1/5 = 7/4 . x + 11/5

3/4 . x - 7/4 . x = 1/5 + 11/5

x (3/4 - 7/4) = 12/5

-x = 12/5

x = -12/5

Vậy x = -12/5.

24 tháng 5 2020

\(M= \dfrac{3^2}{2.5} +\dfrac{3^2}{5.8} +\dfrac{3^2}{8.11}+...+\dfrac{3^2}{98.101}\)

\(M= \) \( \dfrac{9}{2.5} +\dfrac{9}{5.8} +\dfrac{9}{8.11}+...+\dfrac{9}{98.101}\)

\(M=3(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+ \dfrac{3}{98.101})\)

\(M= 3(\dfrac{1}{2} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11})\)

\(M= 3(\dfrac{1}{2}-\dfrac{1}{11})\)

\(M=3(\dfrac{11}{22}- \dfrac{2}{22})\)

\(M=3.\dfrac{9}{22}\)

\(M=\dfrac{27}{22}\)

7 tháng 5 2021

S= 23/2.5+23/5.8+....................+23/53.56

S=8/2.5+8/5.8+8/8.11+.......+8/53.56

S=8/3.(3/2.5+3/5.8+3/8.11+...........+3/53.56)

S=8/3.(1/2-1/5+1/5-1/8+1/8-...........+1/53-1/56)

S=8/3.(1/2-1/56)

S=8/3.27/56

S=9/7

nhớ t ick cho mình nha

11 tháng 8 2015

\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}\right)=3.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=3.\left(\frac{11}{22}-\frac{2}{22}\right)=3.\frac{9}{22}=\frac{27}{22}\)

6 tháng 3 2020

\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)

=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

\(3\left(\frac{1}{2}-\frac{1}{17}\right)\)

=\(\frac{45}{34}\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)

=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)

=3(1/2-1/17)

=45/34

cô Nhung ơi k đúng cho con đi cô pls