`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`

`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`

`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`

`A=1/2xx(1/[1xx2]-1/[99xx100])`

`A=1/2xx(1/2-1/9900)`

`A=1/2xx(4950/9900-1/9900)`

`A=1/2xx4949/9900`

`A=4949/19800`

\(A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}\)

\(A=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right):2\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)

\(A=\dfrac{4949}{9900}:2\)

\(A=\dfrac{4949}{19800}\)

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

C=1x2x3+2x3x4+3x4x5+...+50x51x52
Nhân C với 4 ta được:
C x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+50x51x52x4
C x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 50x51x52x(53-49)
C x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... +49x 50x51x52 - 50x51x52x53
Sau khi cộng - trừ giản ước ta có : C x 4 = 50x51x52x53
C = 50x51x52x53 : 4 = 1756950

\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)

\(=\frac{1}{2}\times\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}+\frac{13-11}{11\times13}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{13}\right)=\frac{6}{13}\)

Do đó ta có: 

\(\frac{6}{13}\times y=\frac{3}{5}\)

\(\Leftrightarrow y=\frac{13}{10}\).

ngu các em học quá ngu

\((\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11})\cdot y=\frac{2}{3}\)

\(\Rightarrow(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11})\cdot y=\frac{2}{3}\)

\(\Rightarrow1-\frac{1}{11}\cdot y=\frac{2}{3}\)

\(\Rightarrow\frac{10}{11}\cdot y=\frac{2}{3}\)

\(\Rightarrow y=\frac{2}{3}:\frac{10}{11}=\frac{11}{15}\)

Vậy :\(y=\frac{11}{15}\)

Bạn có muốn mình thử lại không?

không cảm ơn bạn

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)

=> \(\frac{5}{11}y=\frac{2}{3}\)

=>y = \(\frac{2}{3}:\frac{5}{11}\)

=> y = \(\frac{22}{15}\)

cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm

bạn phạm khánh hà ơi dấu chấm ở giữa các phân số có nghĩa là dấu nhân đó

Giải:

\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}\right)y=-\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{1}{2}.\dfrac{10}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow\dfrac{5}{11}=-\dfrac{2}{3y}\)

\(\Leftrightarrow15y=-22\)

\(\Leftrightarrow y=-\dfrac{22}{15}\)

Vậy ...

ta có : ( 1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 ).y = -2/3

⇒ ( 1- 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 ) . y = -2/3

⇒ ( 1 - 1/11 ) . y = -2/3

⇒ 10/11 . y = -2/3

⇒ y = -2/3 : 10/11

⇒ y = -11/15 .

Vậy y = -11/15

\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(2.\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(2.\frac{10}{11}.y=\frac{2}{3}\)

\(\frac{20}{11}.y=\frac{2}{3}\)

\(\Rightarrow y=\frac{11}{30}\)

Study well