=(3-1)+(7-5)+...+(99-97) (có 50 cặp)

=2+2+..+2 

=2*50

=100

   -1 + 3- 5 + 7-...-97 + 99  ( có 50 số hạng )

= (3 - 1)+(7 - 5)+ ... +(99 -  97)   ( có 25 nhóm )

= 2 + 2 + ... + 2  (có 25 số hạng 2)

= 25.2

=50

a. [ -2/3 + 3/7 ] : 4/5 + [ -1/3 + 4/7 ]  : 4/5 = 0

b. 5/9 : [ 1/11 - 5/22 ] + 5/9 : [ 1/15 - 2/3 ] = -5

HT/nhớ k cho tôi nha

Ai làm xong sớm nhất tui k cho

nntttgfgfgt

-1

Đặt tên cho biểu thức là A

A x 2/3 = 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9+ ... + 2/99x101

Ax2/3 = 3-1/1x3 + 5-3/3x5 + 7-5/5x7 + ... + 101/99

Ax2/3= 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

Ax2/3=1-1/101

Ax2/3=100/101

A=100/101:2/3

A=150/101

tổng trên sẽ là:

3/1x3+....x101=2007

đáp số:2007

k mình nha bạn

Đặt BT trên là A

\(\frac{2}{5}.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

\(\frac{2}{5}.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{100}{101}.\frac{5}{2}=\frac{250}{101}\)

Phải là 100/101 : 2/5 mới đúng chứ

2.08813781251 nhé

Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{5}{2}.\frac{100}{101}=\frac{5.50}{101}=\frac{550}{101}\)

Ta có: \(C=1\cdot3+3\cdot5+5\cdot7+\cdots+97\cdot99\)

\(=1\cdot\left(1+2\right)+3\left(3+2\right)+\cdots+97\left(97+2\right)\)

\(=\left(1^2+3^2+\cdots+97^2\right)+2\left(1+3+\cdots+97\right)\)

Đặt \(A=1^2+3^2+\cdots+97^2\)

=>\(A=1^2+2^2+3^2+4^2+\cdots+98^2-\left(2^2+4^2+\cdots+98^2\right)\)

\(=\left(1^2+2^2+\ldots+98^2\right)-2^2\left(1^2+2^2+\cdots+49^2\right)\)

\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}-4\cdot\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\)

\(=\frac{98\cdot99\cdot197}{6}-\frac{4\cdot49\cdot50\cdot99}{6}=49\cdot33\cdot197-4\cdot49\cdot25\cdot33\)

\(=49\cdot33\cdot\left(197-4\cdot25\right)=49\cdot33\cdot\left(197-100\right)=49\cdot33\cdot97\)

=156849

Đặt \(B=1+3+\cdots+97\)

Số số hạng của dãy số là: \(\frac{97-1}{2}+1=\frac{96}{2}+1=48+1=49\) (số)

Tổng của dãy số là: \(B=\left(97+1\right)\cdot\frac{49}{2}=49\cdot49=2401\)

Ta có: \(C=\left(1^2+3^2+\cdots+97^2\right)+2\left(1+3+\cdots+97\right)\)

=156849+2*2401

=156849+4802

=161651

Ta có: 

\(\dfrac{2}{5}\cdot\dfrac{2}{9}+\dfrac{2}{5}:\dfrac{9}{7}+2\)

\(=\dfrac{2}{5}\cdot\dfrac{2}{9}+\dfrac{2}{5}\cdot\dfrac{7}{9}+2\)

\(=\dfrac{2}{5}\cdot\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+2\)

\(=\dfrac{2}{5}\cdot1+2\)

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

Ta thấy \(\dfrac{2}{5}+2>\dfrac{2}{5}\)

Vậy: \(\dfrac{2}{5}\cdot\dfrac{2}{9}+\dfrac{2}{5}:\dfrac{9}{7}+2>\dfrac{2}{5}\)

_________________________

\(\dfrac{8}{9}:\left(\dfrac{1}{11}-\dfrac{5}{12}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{8}{9}:-\dfrac{43}{132}+\dfrac{5}{9}:-\dfrac{3}{5}\)

\(=\dfrac{8}{9}\cdot-\dfrac{132}{43}+\dfrac{5}{9}\cdot-\dfrac{5}{3}\)

\(=-\dfrac{352}{129}+-\dfrac{25}{27}\)

\(=\dfrac{-4243}{1161}\)