\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+.........+\dfrac{1}{2015}+\dfrac{1...">
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a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)

b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)

c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)

 

18 tháng 4 2020

\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)

\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)

\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)

Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)

3 tháng 5 2018

mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

20 tháng 9 2017

Mấy bài dễ tự làm nhé:D

1)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)

Ta có điều phải chứng minh

16 tháng 6 2019

câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)

b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)

\(6^{x+1}-6^x=6^{14}-6^{13}\)

\(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)

\(x=13\)

c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)

\(2^{x+3}-2^x=2^{13}-2^{10}\)

\(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)

\(x=10\)

d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

\(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)

\(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

\(x=13\)

15 tháng 6 2019

câu d chưa có đóng ngoặc kìa bn

19 tháng 11 2017

8,A=\(\dfrac{9}{10}-\left(\dfrac{1}{10\times9}+\dfrac{1}{9\times8}+\dfrac{1}{8\times7}+...+\dfrac{1}{2\times1}\right)\)

=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+...+\dfrac{1}{2}-1\right)\)

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

=\(\dfrac{9}{10}-\dfrac{\left(-9\right)}{10}\)

=\(\dfrac{9}{5}\)

hihahihahiha

28 tháng 2 2018

bay bị chập p

NV
15 tháng 6 2019

a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)

\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)

\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)

\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)

b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)

\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)

\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)

\(\Rightarrow x=16\)

c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)

\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)

\(\Rightarrow x=20\)

\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)

\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)

\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)

\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)

23 tháng 8 2018

=1/8-1/5+1/7*3/8-3/5+3/7 + 1/2+1/3-1/5*3/4+1/2-3/10

=19/280*57/280+19/30*19/20

=19/280.280/57+19/30.20/19

=1/1.1/3+1/3.2/1

=1/3+2/3=3/3

=1

23 tháng 8 2018

* : chia

. : nhân