Ta có : 

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)

\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)

\(A=\frac{1}{2016}\)

Vậy \(A=\frac{1}{2016}\)

Chúc bạn học tốt ~ 

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)

\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)

\(\Rightarrow A=\frac{1}{2016}\)

k đúng cho mk nha

Đề là gì thế bạn? Tính hay So sánh?

đề là tính các bạn ạ. Mình xin lỗi vì quên ko ghi đề.

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)

\(=\frac{\left(1.2.3...2016\right)\left(3.4.5...2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)

\(=\frac{2018}{2017.2}=\frac{1009}{2017}\)

\(A=\left(-\frac{1}{2}\right).\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)......\left(-\frac{2013}{2014}\right)=\left(-\frac{1}{2014}\right)\) (Do các thừa số đều âm và A có (2014-2)+1=2013 thừa số nên A mang giá trị âm)

\(B=-\frac{1}{2015}\)

=> A<B (|A|>|B|)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)

Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)

          \(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)

          \(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)

          \(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)

          \(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)

Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)

\(A=\frac{1}{102}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)

\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)

\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có : 

\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)

\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)

\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)

\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)

\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)

\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)

\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)

Mà : 

\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)

\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)

Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế ) 

\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)

\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 ) 

\(\Rightarrow\)\(A>3\) ( điều phải chứng minh ) 

Vậy \(A>3\)

Chúc đệ học tốt ~ 

c, 

\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)

vì \(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{6}< \frac{6}{7}\)

.............................

\(\frac{9999}{10000}< \frac{10000}{10001}\)

nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)

\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)

\(\Rightarrow C< \frac{1}{100}\)

bt lm mỗi một câu :v

ae nhiệt tình lên nào

,mình sửa lại đề:

\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)

xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015

=\(\frac{2013}{2013}\)

=\(1\)

vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)

TA CÓ

y=1/2.2/3.3/4..............2013/2014.2014/2015

y=(1.2.3...............2014)/(2.3.4..............2015)

y=1/2015

NHẤT ĐỊNH SẼ CÓ PHÂN SỐ \(1-\frac{2014}{2014}=0\)

NÊN tích dãy số đó là 0

tk nha

100 ngày