\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{1}{3}+\dfrac{1}{25}\)

\(=\dfrac{28}{75}\)

cám ơi bạn

cảm ơn bạn rất nhiều

B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13

   = 1/2 * ( 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13)

   = 1/2 * ( 1/3 - 1/5 + 1/5 -1/7 + ...+ 1/11 - 1/13)

   = 1/2 * ( 1/3 - 1/11)

   = 1/2 * 8/33

   = 4/33

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(B=\frac{1}{2}.\frac{12}{39}\)

\(B=\frac{2}{13}\)

\(\Rightarrow2B=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(\Rightarrow\)\(2B=\frac{1}{3}-\frac{1}{13}\)\(=\frac{13}{39}-\frac{3}{39}=\frac{10}{39}\)\(\Rightarrow B=\frac{10}{\frac{39}{2}}=\frac{10}{39\cdot2}=\frac{5}{39}\)

\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)

\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)

\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

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

\(=6-\frac{12}{13}\)

\(=\frac{66}{13}\)

Ủng hộ mk nha ^_-

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\times\frac{98}{303}\)

\(A=\frac{49}{303}\)

A= \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ ... + \(\frac{1}{9999}\)

A= \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\) + ... + \(\frac{1}{99.101}\)

2. A= \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) + ... + \(\frac{2}{99.101}\)

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

2.A= \(\frac{1}{3}\) - \(\frac{1}{101}\)

2.A= \(\frac{101}{303}\) - \(\frac{3}{303}\)

2.A= \(\frac{98}{303}\)

A  = \(\frac{98}{303}\) : 2

A  = \(\frac{49}{303}\)

Vay A=\(\frac{49}{303}\)

43/303

ket qua la ;49/303

tick nha

Cảm ơn mọi người nhiều!

bài 15 băng 72/13 bạn nha hoc ngu

giải kiểu gì vậy. ghi vào vở như nào

S=1+2+3+...+999

Số số hạg là: (999-1):1+1=999 (số)

Tổng là: (999+1) * 999 :2=499500

S2=21+23+25+...+1001

Số số hạng : (1001-21) : 2 + 1=491 (số)

Tổng là: (1001+21) * 491 :2=250901

S3=15+25+35+...+115

Số số hạng: (115-15) :10+1=11(số)

Tổg là: (115+15)*11:2=715

Hok tốt~

#include <bits/stdc++.h>

using namespace std;

long long s,i;

int main()

{

s=0;

for (i=1; i<=100; i++)

if (i%2==1) s=s+i*i;

cout<<s;

return 0;

}

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn