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7 tháng 5 2017

Lời giải:

a, Đặt \(A=\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(\Rightarrow A=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+...+\dfrac{-1}{9.10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+...-\dfrac{1}{9}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-1}{4}+\dfrac{1}{10}\)

\(\Rightarrow A=\dfrac{-3}{20}\)

7 tháng 5 2017

\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{9.10}\\ =\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)

\(=\dfrac{-1}{4}-\dfrac{-1}{10}\\=\dfrac{-1}{4}+\dfrac{1}{10}\\=\dfrac{-5}{20}+\dfrac{2}{20}\\=\dfrac{-3}{20}\)

AH
Akai Haruma
Giáo viên
31 tháng 12 2018

Lời giải:

a)

\(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+....+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+...+\frac{-1}{9.10}\)

\(=\frac{4-5}{4.5}+\frac{5-6}{5.6}+\frac{6-7}{6.7}+....+\frac{9-10}{9.10}\)

\(=\frac{1}{5}-\frac{1}{4}+\frac{1}{6}-\frac{1}{5}+\frac{1}{7}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{9}\)

\(=\frac{1}{10}-\frac{1}{4}=-\frac{3}{20}\)

b)

\(2B=5+\frac{8}{11}+\frac{3}{11}+\frac{1}{15}+\frac{13}{15.2}\)

\(=5+\frac{11-3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{15-2}{15.2}\)

\(=5+1-\frac{3}{11}+\frac{3}{11}+\frac{1}{15}+\frac{1}{2}-\frac{1}{15}\)

\(=5+1+\frac{1}{2}=\frac{13}{2}\Rightarrow B=\frac{13}{4}\)

31 tháng 12 2018

Thanks

mk làm câu a cách khác ; nhưng cũng = \(\dfrac{-3}{20}\)

27 tháng 3 2017

2. Tính:

a, \(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

=\(\left(\dfrac{-1}{20}+\dfrac{-1}{72}\right)+\left(\dfrac{-1}{30}+\dfrac{-1}{90}\right)+\left(\dfrac{-1}{42}+\dfrac{-1}{56}\right)\)

=\(\left(\dfrac{-18}{360}+\dfrac{-5}{360}\right)+\left(\dfrac{-3}{90}+\dfrac{-1}{90}\right)+\left(\dfrac{-4}{168}+\dfrac{-3}{168}\right)\)

=\(\dfrac{-23}{360}+\dfrac{-4}{90}+\dfrac{-7}{168}\)

=\(\dfrac{-23}{360}+\dfrac{-16}{360}+\dfrac{-15}{360}\)=\(\dfrac{-54}{360}=\dfrac{-3}{20}\)

b, \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

=\(\dfrac{5}{2}+\dfrac{4}{1}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{1}{15}+\dfrac{1}{15}.\dfrac{13}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\left(\dfrac{4}{1}+\dfrac{3}{2}\right)+\dfrac{1}{15}.\left(\dfrac{1}{2}+\dfrac{13}{4}\right)\)

=\(\dfrac{5}{2}+\dfrac{1}{11}.\dfrac{11}{2}+\dfrac{1}{15}.\dfrac{15}{4}\)

=\(\dfrac{5}{2}+\dfrac{1}{2}+\dfrac{1}{4}\)

=\(\dfrac{10}{4}+\dfrac{2}{4}+\dfrac{1}{4}\)

=\(\dfrac{13}{4}\)

27 tháng 3 2017

3. Tìm x

a, \(\dfrac{x-5}{8}=\dfrac{18}{x-5}\)

\(\left(x-5\right).\left(x-5\right)=8.18\)

\(\left(x-5\right)^2=144\)

\(x-5=\sqrt{144}\)

\(x-5=12\)

\(x=12+5\)

\(x=17\)

b,\(\left(x-2\right)^{10}=\left(2-x\right)^8\)

\(x^{10}-2^{10}=x^8-2^8\)

\(x^{10}+x^8=2^{10}+2^8\)

\(\Rightarrow x=2\)

10 tháng 7 2023

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)\(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)\(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

10 tháng 7 2023

Em cảm ơn ạ.

6 tháng 3 2023

\(\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{13}{15\cdot4}\\ =\dfrac{5}{2}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{13}{60}\\ =\dfrac{193}{60}\)

b: \(C=\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\cdot\dfrac{3-2-1}{6}=0\)

27 tháng 5 2022

Quãng sông AB dài là :

    8 giờ 24 phú x 10 = 84 (km)

    Vận tốc cua ca nô khi xuôi dòng là :

    10 + 2 = 12 (km/giờ )

    Thời gian ca nô đi xuôi dòng là :

    84 : 2 = 7 (giờ )

         Đáp số : 7 giờ 

29 tháng 5 2022

= -( 1/1.2 +1/2.3 +1/3.4 + 1/4.5 +1/5.6 + 1/6.7 +1/7.8+ 1/8.9 +1/9.10)

= -( 1-1/2 + 1/2-1/3+1/3 -1/4+.....+1/9-1/10)

= -(1-1/10)

= - 9/10

27 tháng 5 2022

`[-1]/2+[-1]/6+[-1]/12+[-1]/20+[-1]/30+[-1]/42+[-1]/56+[-1]/72+[-1]/90`

`=(-1)(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`

`=(-1)(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)`

`=(-1)(1-1/10)`

`=(-1). 9/10=-9/10`

27 tháng 5 2022

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

A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{2\times3}\)\(\dfrac{-1}{3\times4}\)\(\dfrac{-1}{4\times5}\)\(\dfrac{-1}{5\times6}\)\(\dfrac{-1}{6\times7}\)\(\dfrac{-1}{7\times8}\)\(\dfrac{-1}{8\times9}\)+ + \(\dfrac{-1}{9\times10}\)

A = - (\(\dfrac{1}{2}\)\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)\(\dfrac{1}{4}\)\(\dfrac{1}{5}\)\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\))

A = -(1-\(\dfrac{1}{10}\))

A = \(\dfrac{-9}{10}\)

27 tháng 3 2022

dễ mà 

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{10}{10}-\dfrac{1}{10}=\dfrac{9}{10}\)