456 x 128 / 451 x 128 =58368/57728

123 x 451 / 128 x 451 = 55473/57728

so sánh : 58368/57728 ...>....  55473/ 57728 

vậy suy ra : 456/451 ....>.... 123/128 

tk mk nha mk nhanh nhất

\(\frac{456}{451}\) >    \(\frac{123}{128}\)tích cho mik nhé

Ta có

 \(2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=\frac{2016^{2017}+1}{2016^{2017}+1}+\frac{2015}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)

\(2016B=\frac{2016^{2016}+2016}{2016^{2016}+1}=\frac{2016^{2016}+1}{2016^{2016}+1}+\frac{2015}{2016^{2016}+1}=1+\frac{2015}{2016^{2016}+1}\)

Do \(\frac{2015}{2016^{2017}+1}< \frac{2015}{2016^{2016}+1}\Rightarrow2016A< 2016B\Rightarrow A< B.\)

B = \(\frac{2016^{2015}+1}{2016^{2016}+1}\)< A =\(\frac{2016^{2016}+1}{2016^{2017}+1}\)

Khi a,b cùng dấu thì:

\(\frac{a}{b}\)hoặc \(-\frac{a}{-b}\)\(>0\)

Khi a,b khác dấu:

a dương b âm

\(\frac{a}{-b}< 0\)

a âm b dương

\(-\frac{a}{b}< 0\)

tíc mình nha

Ta có:

A = \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2016}\)

\(=2.\left(\frac{1}{60.63}+\frac{1}{63.66}+...+\frac{1}{117.120}\right)+\frac{2}{2016}\)

\(=2.\frac{1}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2016}\)

\(=\frac{2}{3}.\frac{1}{120}+\frac{2}{2016}\)

\(=\frac{1}{180}+\frac{2}{2016}\)

B = \(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2016}\)

\(=\frac{5}{4}.\frac{1}{80}+\frac{5}{2016}\)

\(=\frac{1}{64}+\frac{5}{2016}\)

Vì \(\frac{1}{64}>\frac{1}{180}\) và \(\frac{5}{2016}>\frac{2}{2016}\) nên B > A

Vậy B > A

Thanks nhiều nhé

May mà đc cậu giúpNguyễn Huy Tú

Tử số của A bằng:

Số số hạng: (2m-2)/2+1=2(m-1)/2+1=m

Tổng= (2+2m).m/2=2(m+1)/2=m.(m+1)

A=m.(m+1)/m=m+1

Câu B tương tự= n+1

Có A<B suy ra m+1<n+1

Suy ra m<n

T..i..c..k mk nha

\(A=\frac{2+4+6+...+2m}{m}=\frac{\left(2m+2\right)\cdot m}{2}=\frac{2\left(m+1\right)\cdot m}{2}=\left(m+1\right)\cdot m\)

\(B=\frac{2+4+6+...+2n}{n}=\frac{\left(2n+2\right)\cdot n}{2}=\frac{2\left(n+1\right)\cdot n}{2}=\left(n+1\right)n\)

Vì A < B

\(\Rightarrow\left(m+1\right)\cdot m<\left(n+1\right)\cdot n\)

=> m < n

\(A=\frac{2016^{2016}-1+3}{2016^{2016}-1};B=\frac{2016^{2016}-3+3}{2016^{2016}-3}\)

\(A=\frac{2016^{2016}-1}{2016^{2016}-1}+\frac{3}{2016^{2016}-1};B=\frac{2016^{2016}-3}{2016^{2016}-3}+\frac{3}{2016^{2016}-3}\)

\(A=1+\frac{3}{2016^{2016}-1};B=1+\frac{3}{2016^{2016}-3}\)

Vì \(\frac{3}{2016^{2016}-1}< \frac{3}{2016^{2016}-3}\)

\(\Rightarrow1+\frac{3}{2016^{2016}-1}< 1+\frac{3}{2016^{2016}-3}\)

\(\Rightarrow A< B\)

\(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)

\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)

Do  \(\frac{3}{2016^{2016}-1}>\frac{3}{2016^{2016}-3}\)

\(\Rightarrow1+\frac{3}{2016^{2016}-1}>1+\frac{3}{2016^{2016}-3}\)

\(\Rightarrow A>B\)

Vậy \(A>B\)

Chúc bạn học tốt !!! 

Bài 1:

ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

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

\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)

\(\Rightarrow B< \frac{1}{4}\)

Bài 2:

ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Học tốt nhé bn !!