Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
a )
\(\frac{-4}{9}.\frac{1}{3}-\frac{4}{9}.\frac{5}{6}+\frac{3}{7}.\frac{4}{9}\)
\(=\frac{4}{9}.\left(-\frac{1}{3}-\frac{5}{6}+\frac{3}{7}\right)\)
\(=\frac{4}{9}.\left(-\frac{14}{42}-\frac{35}{42}+\frac{18}{42}\right)\)
\(=\frac{4}{9}.\frac{-31}{42}\)
\(=-\frac{62}{189}\)
b )
\(\frac{2}{3}:\frac{3}{7}-\frac{2}{3}:\frac{4}{3}+\frac{2}{3}:\frac{1}{21}\)
\(=\frac{2}{3}.\frac{7}{3}-\frac{2}{3}.\frac{3}{4}+\frac{2}{3}.21\)
\(=\frac{14}{9}-\frac{1}{2}+14\)
\(=\frac{28}{18}-\frac{9}{18}+14\)
\(=\frac{19}{18}+14\)
\(=1+14+\frac{1}{18}\)
\(=15\frac{1}{18}\)
c )
\(\left(5\frac{1}{3}+3\frac{2}{3}\right)-4\frac{1}{3}\)
\(=\left(5+3-4\right)+\left(\frac{1}{3}+\frac{2}{3}-\frac{1}{3}\right)\)
\(=4\frac{2}{3}\)
\(=\frac{14}{3}\)
a) \(-\frac{4}{9}\cdot\frac{1}{3}-\frac{4}{9}\cdot\frac{5}{6}+\frac{3}{7}\cdot\frac{4}{9}\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{1}{3}+\left(-\frac{4}{9}\right)\cdot\frac{5}{6}-\left(-\frac{4}{9}\right)\cdot\frac{3}{7}\)
\(=\left(-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{5}{6}-\frac{3}{7}\right)\)
\(=\left(-\frac{4}{9}\right)\cdot\frac{31}{42}=-\frac{62}{189}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
Ta có : A= 3[1 + 1/(1+2) + 1/(1+2+3) +...+1/(1+2+3+..+100)]
Ta thấy: 1/(1+2) = 1/(2.3/2)=2/(2.3)
1/(1+2+3) = 1/(3.4/2)=2/(3.4)
...
1/(1+2+3+...+100)=1/(101.100/2)=2/(101.100)
Suy ra : A= 3[1+ 2/(2.3) + 2/(3.4) +...+ 2/(100.101)]
= 3.2.[1/2 + 1/(2.3) + 1/(3.4) +...+ 1/(100.101)]
= 6.( 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/100- 1/101)
= 6.(1/2 + 1/2-1/101)
= 6. 100/101
= 600/101
Vậy A = 600/101
Chúc bạn học tốt