sửa lại đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(S=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\)

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

\(S=1-\frac{1}{17}=\frac{16}{17}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{14}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)

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

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

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

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

Vậy S = \(\frac{16}{17}\)

*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

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

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

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

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

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

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

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

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

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

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

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

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

= \(\frac{18}{7}\)

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

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

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

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

Sửa lại đề :  \(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.14}+\frac{3}{14.17}\)

\(A=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.14}+\frac{3}{14.17}\right)\)

\(A=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

\(A=\frac{3}{3}\left(1-\frac{1}{17}\right)\)

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

P/S : Ở chỗ 3/11.14 có lẽ bạn ghi sai đề , mình nghĩ là 3/10.14 mới đúng

Sửa lại đề :  A=31.4+34.7+37.10+310.14+314.17A=31.4+34.7+37.10+310.14+314.17

A=3.(11.4+14.7+17.10+110.14+314.17)A=3.(11.4+14.7+17.10+110.14+314.17)

A=3.13(1−14+14−17+17−110+110−114+114−117)A=3.13(1−14+14−17+17−110+110−114+114−117)

A=33(1−117)A=33(1−117)

A=1617A=1617

P/S : Ở chỗ 3/11.14 có lẽ bạn ghi sai đề , mình nghĩ là 3/10.14 mới đúng

Ngu ms đi hỏi câu này

ÓC CHÓ

Bài tập 1:

S=2/15+2/35+2/63+2/99+2/143

\(\Rightarrow\)S=2/3x5 +2/5x 7 +2/7x9 +2/9x11 +2/11x13

\(\Rightarrow\)S=1/3 -1/5 +1/5 - 1/7 +1/7 -1/9 +1/9 -1/11 +1/11 -1/13

\(\Rightarrow\)S=1/3 -1/13

\(\Rightarrow\)S=13/39 -3/39

\(\Rightarrow\)S=10/39

S=3/1.4 +3/4.7+3/7.11 ..........sai đề rồi

Bài 2

A=5/11.16+5/16.21+5/21.26+...+5/61.66

\(\Rightarrow\)A=1/11+1/16+1/16-1/21+1/21-1/26+....+1/61-1/66

\(\Rightarrow\)A=1/11-1/66

\(\Rightarrow\)A=6/66-1/66

\(\Rightarrow\)A=5/66

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

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

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

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

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

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

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

\(B=3.\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)=3\left(\frac{25}{200}-\frac{1}{200}\right)=3\cdot\frac{24}{200}=\frac{72}{200}=\frac{9}{25}\)