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29 tháng 5 2017

a, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{299.300}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

\(=1-\dfrac{1}{300}=\dfrac{299}{300}\)

Vậy \(A=\dfrac{299}{300}\)

b, \(B=\dfrac{10^2}{16.26}+\dfrac{10^2}{26.36}+...+\dfrac{10^2}{86.96}\)

\(=10\left(\dfrac{10}{16.26}+\dfrac{10}{26.36}+...+\dfrac{10}{86.96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+...+\dfrac{1}{86}-\dfrac{1}{96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{96}\right)\)

\(=10.\dfrac{5}{96}=\dfrac{25}{48}\)

Vậy...

29 tháng 5 2017

a,\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{299.300}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(A=\dfrac{1}{1}-\dfrac{1}{300}=\dfrac{299}{300}\)

4 tháng 6 2018

Câu b, B=\(\dfrac{5}{1\cdot2}+\dfrac{13}{2\cdot3}+\dfrac{25}{3\cdot4}+...+\dfrac{181}{9\cdot10}\)

\(=\left(\dfrac{1}{1\cdot2}+\dfrac{4}{1\cdot2}\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{12}{2\cdot3}\right)+\left(\dfrac{1}{3\cdot4}+\dfrac{24}{3\cdot4}\right)+...+\left(\dfrac{1}{9\cdot10}+\dfrac{180}{9\cdot10}\right)\)=\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)+\left(\dfrac{4}{1\cdot2}+\dfrac{12}{2\cdot3}+...+\dfrac{180}{9\cdot10}\right)\)

=\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)\(+\left(2+2+2+.......+2\right)\)

=\(\dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-......-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\dfrac{1}{10}+\left(2\cdot9\right)\)

=\(1-\dfrac{1}{10}+18\) \(=\dfrac{9}{10}+18\)

=18.9

4 tháng 6 2018

a, \(\dfrac{\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}}{\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}}=\dfrac{\dfrac{15}{10}-\dfrac{4}{10}+\dfrac{1}{10}}{\dfrac{18}{12}-\dfrac{8}{12}+\dfrac{1}{12}}=\dfrac{\dfrac{15-4+1}{10}}{\dfrac{18-8+1}{12}}=\dfrac{\dfrac{12}{10}}{\dfrac{11}{12}}=\dfrac{72}{55}\)

Bài 1: 

a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)

\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)

\(=\dfrac{1}{2}\)

c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)

\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

9 tháng 1 2021

thanks 

1 tháng 11 2023

a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

1 tháng 11 2023

cứu 

3 tháng 10 2021

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

3 tháng 10 2021

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

19 tháng 10 2021

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)

AH
Akai Haruma
Giáo viên
3 tháng 12 2017

Lời giải:

Ta có:

\(\frac{1}{1.2^2}=\frac{1}{2^2}\)

\(2.3^2>3^2\Rightarrow \frac{1}{2.3^2}< \frac{1}{3^2}\)

\(3.4^2> 4^2\Rightarrow \frac{1}{3.4^2}< \frac{1}{4^2}\)

...........

\(49.50^2> 50^2\Rightarrow \frac{1}{49.50^2}< \frac{1}{50^2}\)

Cộng theo từng vế các BĐT:

\(\Rightarrow \frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+....+\frac{1}{49.50^2}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)

\(\Leftrightarrow A< B\)

Vậy ta có đpcm.

16 tháng 3 2017

Ta có: \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{10^2}{10.11}\)

\(=\dfrac{2.2.3.3...10.10}{2.2.3.3.4...10.11}\)

\(=\dfrac{1}{11}\)

Vậy tích trên có giá trị \(=11.\)

9 tháng 5 2022

999/1000(hình như v)

9 tháng 5 2022

Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:

\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)