mk đi thi hsg mk bik chắc chắn mk ko sai đâu k mk nhé
3/1.4 +3/4.7+....+3/14.17
=1/1-1/4+1/4-1/7+...+1/14-1/17
=1-1/17
=16/17

\(\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...\)\(\frac{3}{14}-\frac{3}{17}\)

\(\frac{3}{1}-\frac{3}{17}\)

= \(\frac{18}{7}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(=1-\frac{1}{17}\)

\(=\frac{16}{17}\)

mình nghĩ bài náy sai đề rồi

*S=1-1/4+1/4-1/7+1/7-1/11+1/11-1/14+1/14-1/17

S=1-1/17=16/17

*M=2(1/1.2+1/2.3+...+1/15.16)

M=2(1-1/2+1/2-1/3+..+1/15-1/16)

M=2(1-1/16)

M=2.15/16

M=15/8

:w

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(S=1-\frac{1}{17}\)

\(S=\frac{16}{17}\)

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{15.16}\)

\(M=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(M=2.\left(1-\frac{1}{16}\right)\)

\(M=2.\frac{15}{16}\)

\(=\frac{30}{16}=\frac{15}{8}\)

lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé

a) \(\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}\)

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

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

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

b) \(\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}\)

\(=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\times\frac{3}{25}=\frac{9}{25}\)

Ta có \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)

\(\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)

\(\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow3.\left(1-\frac{1}{200}\right)\)

\(\Rightarrow3.\frac{199}{200}=\frac{597}{200}\)

a) \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

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

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

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

\(=\frac{250}{101}\)

Ta có : \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{89}{270}\)

\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{n\left(n+3\right)}=\frac{267}{270}\)

\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}=\frac{267}{270}\)

\(\Rightarrow1-\frac{1}{n+3}=\frac{267}{270}\)

=> \(\frac{1}{n+3}=\frac{1}{90}\)

=> n + 3 = 90

=> n = 87 

Nhân cả 2 vế với 3 ta được:

\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{n\left(n+3\right)}=\frac{89}{90}.\)

Vậy tử số của các phân số trên đã bằng hiệu của 2 thừa số ở mẫu số.(Ngoại trừ P/S\(\frac{89}{90}.\))

=> ta được:

\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{n}-\frac{1}{n+3}=\frac{89}{90}.\)

Rút gọn hết ta được :

\(1-\frac{1}{n+3}=\frac{89}{90}\)

\(\frac{1}{n+3}=1-\frac{89}{90}\)

\(\frac{1}{n+3}=\frac{1}{90}.\)

Vì 1=1 => n+3=90

          n = 90-3

          n=87

Vậy n=87.

                                                                    Đ/S:87

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.11}+....+\frac{1}{n\left(n+3\right)}=\frac{89}{270}\)

\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+3}\right)=\frac{89}{270}\)

\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{n+3}\right)=\frac{89}{270}\div\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{n+3}=\frac{89}{90}\)

\(\Leftrightarrow\frac{1}{n+3}=\frac{1}{1}-\frac{89}{90}=\frac{1}{90}\)

\(\Leftrightarrow n+3=90\Rightarrow n=90-3=87\)

tớ ko biết

S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

S=3/1.4+3/4.7+3/7.10+.....+3/97.100

S=1/1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100

S=1-1/100

S=99/100

Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(A=\frac{1}{2}-\frac{1}{17}\)

\(A=\frac{15}{34}\)

= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)

\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)

\(=3\cdot\frac{23}{200}\)

đúng

Đặt 3 ra ngoài

\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)

Vậy \(B=\frac{9}{25}\)

Chúc bn học tốt..!