Question

So sánh A= \(\frac{3}{1.3}\)+ \(\frac{3}{3.5}\)+ \(\frac{3}{5.7}\)+...+\(\frac{3}{101.103}\) với  1

Answer 1

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{101.103}\)

\(=>A=\frac{3}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\right)\)

\(=>A=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)

\(=>A=\frac{3}{2}.\left(1-\frac{1}{103}\right)=\frac{3}{2}.\frac{102}{103}=\frac{153}{103}>1\) (vì 153>103)

Vậy A>1

Answer 2

sorry,dòng thứ 2 sửa lại:\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{3}{101.103}\right)\) nhé!

Answer 3

Bạn xem lời giải của mình nhé:

Giải:

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{101.103}\\ =\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\\ =\frac{3}{2}.\left(1-\frac{1}{103}\right)\\ =\frac{3}{2}.\frac{103-1}{103}=\frac{3}{2}.\frac{102}{103}=\frac{153}{103}=1\frac{50}{103}\)

Chúc bạn học tốt!

Answer 4

2/3 A = 2/3. ( 3/1.3 + 3/3.5 + 3/5.7 +...+ 3/101.103 )

2/3 A= 2/1.3 + 2/3.5 + 2/5.7 +...+2/101.103

2/3 A= 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - .... - 1/101 + 1/101 - 1/103
2/3 A= 1/1 - 1/103 
2/3 A= 102/103
       A= 102/103 : 2/3
       A= 153/103
vì 143/103 > 1 => A >1
 

Answer 5

với 1

Answer 6

đến đấy thì bạn tự so sánh nha!

Answer 7

Mình không ghi ra phân số như \(\frac{2}{3}\) nhé ( lười )

Answer 8

thanks all of you

Answer 9

A=3/1.3+3/3.5+...+3/101.103

A=3/2.(2/1.3+2/3.5+...+2/101.103)

A=3/2.(1-1/3+1/3-1/5+...+1/101-1/103)

A=3/2.(1-1/103)

A=3/2.102/103

A=153/103

Vì 153/103>1  =>    A>1

 

Answer 10

\(\frac{3}{3.1}=1\Rightarrow A=\frac{3}{3.1}+.......>1.\)

 

Answer 11

\(A=\frac{3}{2}\cdot\frac{3}{1\cdot3}+\frac{3}{2}\cdot\frac{3}{3\cdot5}+\frac{3}{2}\cdot\frac{3}{5\cdot7}+...+\frac{3}{2}\cdot\frac{3}{101\cdot103}\)với 1

A=\(\frac{3}{2}\)(\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+.......+\frac{3}{101\cdot103}\))A=\(\frac{3}{2}\)(\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\))A=\(\frac{3}{2}\)(\(\frac{1}{1}-\frac{1}{103}\))A=\(\frac{3}{2}\)(\(\frac{103}{103}-\frac{1}{103}\))A=\(\frac{3}{2}\)\(\frac{102}{103}\)A=\(\frac{306}{206}\)

\(\Rightarrow\frac{306}{206}>1\)

Answer 12

\(A=3\left(\frac{1}{1}-\frac{1}{3}\right)\frac{1}{3}+3\left(\frac{1}{3}-\frac{1}{5}\right)\frac{1}{2}+......+3\left(\frac{1}{101}-\frac{1}{103}\right)\frac{1}{2}\)

\(A=3x\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{103}\right)\)

\(A=\frac{3}{2}x\frac{102}{103}\)

\(A=\frac{153}{103}\)

\(A=1\frac{50}{103}\)

 \(1\frac{50}{103}\) > 1

-> A >1

 

Answer 13

chắc chắn là A> 1 rồi

lớn hơn nha(A>1)

Answer 14

Mấy chế dài dòng quá, noi gương cách ngắn nhất đây này!

\(A=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{101.103}\)

\(=\frac{3}{3}+\frac{3}{3.5}+...+\frac{3}{101.103}\)

\(=1+\frac{3}{3.5}+...+\frac{3}{101.103}>\)\(1\)