65465412

a) \(16+\left(27-7\cdot6\right)-\left(94\cdot7-27\cdot99\right)\)

\(=16+27-7\cdot6-94\cdot7+27\cdot99\)

\(=16+27\left(1+99\right)-7\left(6+94\right)=16+2700-700=2016\)

b)\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)

\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+\frac{2}{7}-\frac{2}{10}+...+\frac{2}{97}-\frac{2}{100}\right)\)

\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\cdot\frac{99}{50}=\frac{33}{50}\)

vô lí! đoạn cuối phải là 2/196.199 chứ

Sai đề : \(\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)

\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)

Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)

\(B=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)

\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)

\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)

Chúc bạn học tốt !!!

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

Có : 3/5 A = 3/1.4 + 3/4.7 + 3/7.10 + ..... + 3/307.310

= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ........ + 1/307 - 1/310

= 1 - 1/310

= 309/310

=> A = 309/310 : 3/5 = 103/62

Tk mk nha

        \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{307.310}\)

\(=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{307.310}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{307}-\frac{1}{310}\right)\)

\(=\frac{5}{3}\left(1-\frac{1}{310}\right)\)

\(=\frac{5}{3}.\frac{309}{310}=\frac{103}{62}\)

A=(5/1-5/4+5/4-5/7+...+5/307-5/310):3

A=(5/1-5/310):3

A=103/62

=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{28}-\frac{2}{31}\right)\)

=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{31}\right)=\frac{20}{31}\)

Bấm đúng cho tui, đi mà. CHÚC BẠN HỌC GIỎI

bài giải đó là sai giả như vầy nè

\(=\frac{1}{3}\cdot2\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{31}\right)\)

\(=\frac{2}{3}\left(1-\frac{1}{31}\right)\)

=\(\frac{2}{3}\cdot\frac{30}{31}=\frac{20}{31}\)