A = \(\dfrac{7}{3\times6}\) + \(\dfrac{7}{6\times9}\) + \(\dfrac{7}{9\times12}\) + \(\dfrac{7}{12\times15}\)+ .....+\(\dfrac{7}{96\times99}\)

A = \(\dfrac{7}{3}\) x ( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+ \(\dfrac{3}{9\times12}\)+ \(\dfrac{3}{12\times15}\)+......+\(\dfrac{3}{96\times99}\))

A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{12}\)+ \(\dfrac{1}{12}\) - \(\dfrac{1}{15}\)+....+ \(\dfrac{1}{96}\) - \(\dfrac{1}{99}\))

A = \(\dfrac{7}{3}\) x ( \(\dfrac{1}{3}\)- \(\dfrac{1}{99}\))

A = \(\dfrac{224}{297}\)

giup minh voi 

Ta có: \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(\Rightarrow B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^6}\)

\(\Rightarrow B=\frac{1-\frac{1}{3^6}}{2}\)

\(A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{93.96}+\frac{3}{96.99}\)

\(A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{93}-\frac{1}{96}+\frac{1}{96}-\frac{1}{99}\)

\(A=1-\frac{1}{99}=\frac{98}{99}\)

Vậy A=\(\frac{98}{99}\)

\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

\(3B=\)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)

\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)

\(3B=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)

\(B=\frac{24}{49}:3=\frac{8}{49}\)

Vậy B=\(\frac{8}{49}\)

Dấu "." là dấu nhân.

_Học tốt_

gcjjjjjjjjjjjhm.f

Vũ Thị Trang lại là một nạn nhân tiếp tục bị báo cáo

nhân cả vế với 3 ta có

Ax3=\(\frac{3}{3x6}\)+\(\frac{3}{6x9}\)+.........+\(\frac{3}{99x102}\)

Ax3=\(\frac{1}{3}\)-\(\frac{1}{6}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{102}\)

Ax=\(\frac{1}{3}\)-\(\frac{1}{102}\)

Ax3=\(\frac{11}{34}\)

A=\(\frac{11}{34}\):3

A=\(\frac{11}{102}\)

gạch đi các số lặp lại thì còn phân số 1/3 và 1/102 lấy \(\frac{1}{3}-\frac{1}{102}=\frac{33}{102}\)

\(A=\frac{1}{3x6}+\frac{1}{6x9}+\frac{1}{9x12}+....+\frac{1}{99x102}\)

\(Ax3=\frac{3}{3x6}+\frac{3}{6x9}+.............+\frac{3}{99x102}\)

\(Ax3=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+.....+\frac{1}{99}-\frac{1}{102}\)

\(Ax3=\frac{1}{3}-\frac{1}{102}\)

\(Ax3=\frac{11}{34}\)

\(A=\frac{11}{34}:3=\frac{11}{102}\)

Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)

Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có : 

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)

\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)

\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)

\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+\frac{7}{13.17}+\frac{7}{17.21}\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\left(\frac{1}{1}-\frac{1}{21}\right)\)

\(=\frac{7}{4}\cdot\frac{20}{21}=\frac{5}{3}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)

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

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

Sủa đề : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)

\(\frac{14}{45}\)thi phai

So sánh A và B biết
B= (1.2+2.4+3.6+4.8+5.10) / (3.4+6.8+9.12+12.16+15.20)
A=111111/666665
ta có:
Tử B=1.2+2.4+3.6+4.8+5.10
=2+8+18+48+50=126
Mẫu B=3.4+6.8+9.12+12.16+15.20
=12+48+108+192+300=660
Phân số 126/660
So sánh phân sô126/660 và phân số 111111/666665
hay so sánh 126.666665 và 111111.660
hay 83999790 và 73333260
vậy 83999790>73333260
nên phân số B>A

\(\text{1x2 + 2x4 + 3x6 + 4x8 + 5x10 / 3x4 + 6x8 + 9x12 + 12x16 + 15x20}\)

\(\text{Mẫu số : 3x4 + 6x8 + 9x12 + 12x16 + 15x20 = 3 x 2 x (1x2 + 2x4 + 3x6 + 4x8 + 5x10)}\)

\(\text{Vậy (1x2 + 2x4 + 3x6 + 4x8 + 5x10) / (3 x 2 x (1x2 + 2x4 + 3x6 + 4x8 + 5x10)) = 1/3x2 = 1/6}\)

\(A=\frac{111111}{666665}=\)bạn tự rút gọn nha !

1x2 + 2x4 + 3x6 + 4x8 + 5x10 / 3x4 + 6x8 + 9x12 + 12x16 + 15x20

Mẫu số : 3x4 + 6x8 + 9x12 + 12x16 + 15x20 = 3 x 2 x (1x2 + 2x4 + 3x6 + 4x8 + 5x10)

Vậy (1x2 + 2x4 + 3x6 + 4x8 + 5x10) / (3 x 2 x (1x2 + 2x4 + 3x6 + 4x8 + 5x10)) = 1/3x2 = 1/6

=> A ... B