`2/(1.4) + 2/(4.7) + 2/(7.10) + 2/(10.13)`

`= 1/3.2(3/(1.4) + 3(4.7) + 3/(7.10) + 3/(10.13))`

`= 2/3 . (1 - 1/4 + 1/4 -1/7 + 1/7 - 1/10 + 1/10 - 1/13)`

`= 2/3 .(1 - 1/13)`

`= 2/3 . 12/13`

`= 8/13`

B=2(1/1.4 +1/4.7+....+1/97.100)

3B= 2(3/1.4+3/4.7+...+3/97.100)

3B=2(1-1/4+1/4-1/7+...+1/97-1/100)

3B= 2(1-1/100)

3B= 2.99/100

3B= 99/50

B=33/50.

B=2/1.4+2/4.7+2/7.10+...+2/97.100

B=2/3.3/1.4+2/3.3/4.7+2/3.3/7.10+...+2/3.3/97.100

B=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100) (dùng phương pháp khử)

B=2/3(1-1/100)

B=2/3.99/100

B=33/50

B=\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)

B=\(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{100}\)

B=\(\frac{2}{1}-\frac{2}{100}=\frac{200}{100}-\frac{2}{100}\)

B=\(\frac{198}{100}=\frac{46}{25}\)

Đặt \(A=2+2^2+2^3+...+2^{15}\)

\(\Rightarrow2A=2^2+2^3+2^4...+2^{15}+2^{16}\)

\(\Rightarrow2A-A=\left(2^2+2^3+2^4+...+2^{16}\right)-\left(2+2^2+2^3+...+2^{15}\right)\)

\(\Rightarrow2A-A=A=2^{16}-2\)

Vậy \(2+2^2+2^3+...+2^{15}=2^{16}-2\)

2+2² +2³+2⁴+......+2¹⁵

Gọi tên biểu thức trên là A

A=2+2² +2³+2⁴+......+2¹⁵

2.A=2.(2+2² +2³+2⁴+......+2¹⁵)

2.A=2² +2³+2⁴+......+2¹⁵+2¹⁶

2.A-A=(2² +2³+2⁴+......+2¹⁵+2¹⁶)-(2+2² +2³+2⁴+......+2¹⁵)

A=2² +2³+2⁴+......+2¹⁵+2¹⁶-2-2²-2³-2⁴+......-2¹⁵

A=2¹⁶-2

A=65534

                                                                  Bài giải

Đặt \(A=2+2^2+2^3+...+2^{15}\)

\(2A=2^2+2^3+2^4+...+2^{16}\)

\(2A-A=2^{16}-2\)

\(A=2^{16}-2\)

A= 2/1.4+2/4.7+2/7.10+...+2/97.100

= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)

= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

= 2.(1/1-1/100)

= 2.(99/100)

=99/50

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

\(A=\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{97\cdot100}\right)\)

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

\(A=\dfrac{2}{3}\cdot\left(1-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}\cdot\dfrac{99}{100}\)

\(A=\dfrac{33}{50}\)

\(A=\dfrac{2}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

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

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

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

Mình mới học lớp 5 , xin lỗi nhé, mình cũng rất muốn giúp bạn nhưng ko đc.

nếu không làm được thì thôi, mong bạn đừng nhắn lời xin lỗi ạ. Không ai như bạn đâu!

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

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

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

\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{2}{3}.\frac{99}{100}\)

\(A=\frac{33}{50}\)