K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

25 tháng 9 2021

d ) 32 . 53 + 92

= 9 . 125 + 81

= 1125 + 81

= 1206

25 tháng 5 2018

ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)

mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)

\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)

\(\Rightarrow A< B\)

25 tháng 5 2018

Câu hỏi của nguyen van nam - Toán lớp 6 - Học toán với OnlineMath

11 tháng 8 2016

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

\(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

11 tháng 8 2016

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

những câu sau cũng áp dụng như vậy nhé

c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)

\(=\dfrac{255}{103}\)

6 tháng 8 2016

1)

a. \(\left(3x^2-50\right)^2=5^4\)

\(\Leftrightarrow3x^4-50=625\)

\(\Leftrightarrow3x^4=675\)

\(\Leftrightarrow x^4=225\)

\(\Leftrightarrow x=\sqrt{15}\) 

2)

a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)

\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)

b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)

\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)

Chọn A

27 tháng 4 2016

A=1+5+5^2+..+5^9/1+5+5^2+...+5^8

=1+5^9/1+5+5^2+...+5^8 

B=1+3+3^2+..+3^9/1+3+3^2+..+3^8

=1+3^9/1+3+3^2+..+3^8

đặt A' =1+5+5^2+...+5^8

5A'=5+5^2+5^3+...+5^9

5A'-A'=5+5^2+5^3+...+5^9-5-1-5-5^2-...-5^8

4A'=5^9-1=>A'=(5^9-1):4

tương tự B'=(3^9-1):4

A=1+5^9/(5^9-1)/4=4.5^9/5^9-1

B=1+3^9/(3^9-1)/4=4.3^9/3^9-1

=> A<B

`@` `\text {Ans}`

`\downarrow`

`a.`

`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`

`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`

`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `

`= 1 - 1 + 1`

`= 1`

`b.`

`B=0,75+2/5+(1/9-1 1/2+5/4)`

`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`

`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`

`= 2 + 1/9 - 11/10`

`= 19/9 - 11/10`

`= 91/90`

`c.`

`(-5/9).3/11+(-13/18).3/11`

`= 3/11*[(-5/9) + (-13/18)]`

`= 3/11*(-23/18)`

`= -23/66`

`d.`

`(-2/3).3/11+(-16/9).3/11`

`= 3/11* [(-2/3) + (-16/9)]`

`= 3/11*(-22/9)`

`= -2/3`

`e.`

`(-1/4).(-2/13)-7/24.(-2/13)`

`= (-2/13)*(-1/4-7/24)`

`= (-2/13)*(-13/24)`

`= 1/12`

`f.`

`(-1/27).3/7+(5/9).(-3/7)`

`= 3/7*(-1/27 - 5/9)`

`= 3/7*(-16/27)`

`= -16/63`

`g.`

`(-1/5+3/7):2/11+(-4/5+4/7):2/11`

`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`

`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`

`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`

`= (-1+1) \div 2/11`

`= 0 \div 2/11 = 0`