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25 tháng 6 2017

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

\(=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{6-7}{6\cdot7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

17 tháng 7 2017

=6/7

nhớ tk cho tui nha ^_^

tk đi tk lại cho

17 tháng 4 2023

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)

25 tháng 3 2022

\(y-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}=1\)

\(y-\dfrac{6}{7}=1\)

\(y=\dfrac{13}{7}\)

2 tháng 7 2016

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)

\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)

2 tháng 7 2016

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)

Vậy A=-79/90

Ta có : \(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-.....-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}=\frac{1}{90}-\frac{80}{90}=\frac{-79}{90}\)

8 tháng 6 2018

Đặt \(A=\left(...\right)\) ( tự ghi ) 

Ta có : 

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}-\frac{1}{9.10}\)

\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(-\frac{1}{9}+\frac{1}{10}\)

\(-A=1-\frac{1}{9}-\frac{1}{9}+\frac{1}{10}\)

\(-A=\frac{79}{90}\)

\(A=\frac{-79}{90}\)

Chúc bạn học tốt ~ 

27 tháng 10 2017

M  =  \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

M  =  \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

M     =   1   -\(\frac{1}{9}\)=\(\frac{8}{9}\)

27 tháng 10 2017

trả lời: 8/9

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}\)

\(=\frac{-79}{90}\)

25 tháng 8 2016

1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2

= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)

= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)

= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)

= 1/90 - (1 - 1/9)

= 1/90 - 8/9

= 1/90 - 80/90

= -79/90

Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

Giải:

1/12+1/20+1/30+...+1/9702

=1/3.4+1/4.5+1/5.6+...+1/98.99

=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99

=1/3-1/99

=32/99

Chúc bạn học tốt!

20 tháng 1 2023

Từ GT ; ta có :  \(\left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right).\dfrac{2}{9}=224\)

\(\Rightarrow\left(x-1\right)=1008\)

\(\Rightarrow x=1009\)

Vậy ... 

27 tháng 1 2023

c= - 1/10 nha