bạn nào trả lời được k cho trong hôm nay vì mk đang cần gấp( giaỉ ra)

Mình nghĩ nên sửa lại chỗ 9 thành 8 hoặc 16 thành 18

\(B=\frac{1.2.6+2.4.12}{1.6.9+2.12.18}=\frac{1.2.6.\left(1+2^3\right)}{1.6.9.\left(1+2^3\right)}\)\(=\frac{1.2.6}{1.6.9}=\frac{2}{9}\)

hoặc 

\(B=\frac{1.2.6+2.4.12}{1.6.8+2.12.16}=\frac{1.2.6.\left(1+2^3\right)}{1.6.8.\left(1+2^3\right)}\)\(=\frac{1.2.6}{1.6.8}=\frac{2}{8}=\frac{1}{4}\)

bạn ấn vào đúng 0 sẽ ra đáp án mình giải 

\(A=\left(1-\dfrac{1}{2014}\right).\left(1-\dfrac{2}{2014}\right).....\left(1-\dfrac{2014}{2014}\right).\left(1-\dfrac{2015}{2014}\right)\)\(=\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{2}{2014}\right).....0.\left(1-\dfrac{2015}{2014}\right)\)

= 0

Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!

Đoàn Đức Hiếu, ... giúp với!

Ta có :

B = \(\dfrac{2015}{1}+\dfrac{2014}{2}+\dfrac{2013}{3}+...+\dfrac{2}{2014}+\dfrac{1}{2015}\) => B = \(\left(1+\dfrac{2014}{2}\right)+\left(1+\dfrac{2013}{3}\right)+...+\left(1+\dfrac{2}{2014}\right)+\left(1+\dfrac{1}{2015}\right)+1\) => B = \(\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2014}+\dfrac{2016}{2015}+\dfrac{2016}{2016}\) => B = \(2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\) Ta có :

\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)}\)

=> \(\dfrac{A}{B}=\dfrac{1}{2016}\)

Vậy \(\dfrac{A}{B}=\dfrac{1}{2016}\)

Ta có:

\(A=\dfrac{9}{a^{2013}}+\dfrac{7}{a^{2014}}\)

\(=\left(\dfrac{8}{a^{2013}}+\dfrac{1}{a^{2013}}\right)+\left(\dfrac{8}{a^{2014}}-\dfrac{1}{a^{2014}}\right)\)

\(=\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)\)

\(B=\dfrac{8}{a^{2014}}+\dfrac{8}{a^{2013}}\)

\(=\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)

Vì \(\dfrac{1}{a^{2013}}>\dfrac{1}{a^{2014}}\Rightarrow\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}>0\)

\(\Rightarrow\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)>\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)

Vậy \(A>B\)