Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
uh.cậu là fan của bts hả.mình cũng thế,nhưng mình thích red velvet hơn
Biến đổi VT ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
\(\Rightarrowđpcm\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1.\left(1-\frac{1}{100}\right)\)
\(=1.\frac{99}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Hình như là fan BTS ms đúng mà~
a)\(\frac{32}{64}-\frac{16}{64}+\frac{8}{64}-\frac{4}{64}+\frac{2}{64}-\frac{1}{64}\le\frac{1}{3}\)
\(\Rightarrow\frac{32-16+8-4+2-1}{64}=\frac{23}{64}\)\
\(\Rightarrow\frac{23}{64}=0,359375;\frac{1}{3}=0,33333...\)
đề sao lạ vậy
1/1+2 + 1/+1+2+3 + ... + 1/1+2+3+...+2014
= 1/(1+2).2:2 + 1/(1+3).3:2 + ... + 1/(1 + 2014).2014:2
= 2/2.3 + 2/3.4 + ... + 2/2014.2015
= 2.(1/2.3 + 1/3.4 + ... + 1/2014.2015)
= 2.(1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015)
= 2.(1/2 - 1/2015)
= 2.1/2 - 2.1/2015
= 1 - 2/2015
= 2013/2015
1/1+2 + 1/+1+2+3 + ... + 1/1+2+3+...+2014
= 1/(1+2).2:2 + 1/(1+3).3:2 + ... + 1/(1 + 2014).2014:2
= 2/2.3 + 2/3.4 + ... + 2/2014.2015
= 2.(1/2.3 + 1/3.4 + ... + 1/2014.2015)
= 2.(1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015)
= 2.(1/2 - 1/2015)
= 2.1/2 - 2.1/2015
= 1 - 2/2015
= 2013/2015
\(P=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1+\frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(P< 1+\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}=\frac{7}{4}-\frac{1}{2019}< \frac{7}{4}\)
Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow S< 1-\frac{1}{100}< 1\Rightarrow S< 1\)
Làm vui đó chủ yếu là nghe link gửi
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}< 1\)
\(A< 1\left(đpcm\right)\)