a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)

b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)

\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)

Bài 1:

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}=\frac{49}{50}\)

Bài 2:

Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)

Vậy A < 2

Bài 3:

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

Bài 4:

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)

A=1-1/2+1/2-1/3+.............................1/49-1/50

A=1-1/50

A=49/50

Tu lam di 

Sao hoi quai day

Phai biet suy nghi cho

bạn tìm trong nâng cao phát triển toán 9 tập 1 ấy nó có ở đấy

đề thiếu bn ơi

phải là \(\frac{1}{99^2}-1\)

A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{98^2}-1\right)\).\(\left(\frac{1}{99^2}-1\right)\)

do tích A có: (99-2)+1=98 thừa số nguyên âm nên tích A dương

A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)=\(\frac{1.3.2.4.3.5...97.99.98.100}{2^2.3^2.4^2...98^2.99^2}\)

=\(\frac{1.2.3.4...98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}\)=\(\frac{1}{99}.\frac{100}{2}\)=\(\frac{50}{99}\)

vậy A=\(\frac{50}{99}\)

#HỌC TỐT#

Mình giúp phần a thôi, phần b chir là áp dụng không có gì khó cả.

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\left(a+b+c=0\right)\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\left(đpcm\right)\)

b, \(A=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{399^2}+\frac{1}{400^2}}\)

\(A=\sqrt{\frac{1}{1^2}+\frac{1}{1^2}+\frac{1}{\left(-2\right)^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{399^2}+\frac{1}{\left(-400\right)^2}}\)

có 1 + 1 - 2 = 1 + 2 - 3 = ... + 1 + 399 - 400 = 0

nên theo câu a ta có : 

\(A=\left|1+\frac{1}{1}-\frac{1}{2}\right|+\left|1+\frac{1}{2}-\frac{1}{3}\right|+...+\left|1+\frac{1}{399}-\frac{1}{400}\right|\)

A = 1 + 1 -1/2 + 1 + 1/2 - 1/3 + 1 + 1/3 - 1/4 + ... + 1 + 1/399 - 1/400

= 400  1/400

= 159999/400

a: \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{99\cdot101}\)

\(=1-\frac13+\frac13-\frac15+\cdots+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

b: \(B=\frac12-\left(\frac{1}{5\cdot11}+\frac{1}{11\cdot17}+\frac{1}{17\cdot23}+\frac{1}{23\cdot29}+\frac{1}{29\cdot35}\right)\)

\(=\frac12-\frac16\left(\frac{6}{5\cdot11}+\frac{6}{11\cdot17}+\frac{6}{17\cdot23}+\frac{6}{23\cdot29}+\frac{6}{29\cdot35}\right)\)

\(=\frac12-\frac16\left(\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{17}+\cdots+\frac{1}{29}-\frac{1}{35}\right)\)

\(=\frac12-\frac16\left(\frac15-\frac{1}{35}\right)=\frac12-\frac16\cdot\frac{6}{35}=\frac12-\frac{1}{35}=\frac{33}{70}\)

mk ko bt 123