Bài 1 .
a) A = 1 + \(\dfrac{1}{2+1}\) + \(\dfrac{1}{2^2+1}\) +\(\dfrac{1}{2^4+1}\) + .....+\(\dfrac{1}{2^{2n}+1}\)

b) B = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) + ... + \(\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)

c) C = \(\dfrac{1.2!}{2}\) + \(\dfrac{2.3!}{2^2}\) +... + \(\dfrac{n.\left(n+1\right)!}{2^n}\) (k! = 1. 2 . 3 ... k)

a, T= (1-\(\dfrac{1}{2}\))+(1-\(\dfrac{1}{4}\))+(1-\(\dfrac{1}{8}\))+...+(1-\(\dfrac{1}{512}\))+(1-\(\dfrac{1}{1024}\))+(1-\(\dfrac{1}{2048}\))+(1-\(\dfrac{1}{4096}\))

b, 4*5¹⁰⁰*(\(\dfrac{1}{5}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{100}}\))+1

Đề:

Cho biết abc = 1. Chứng minh rằng:\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\) là hằng số.

Giải:

Thay 1 = abc vào biểu thức trên, ta có:

\(\frac{a}{ab+a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+abc}\)

\(=\frac{a}{a\left(b+1+ab\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{c\left(a+1+ab\right)}\)

\(=\frac{1}{b+1+ab}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab}\)

\(=\frac{abc}{b+abc+ab}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab}\)

\(=\frac{abc}{b\left(1+ac+a\right)}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab}\)

\(=\frac{ac}{1+ac+a}+\frac{1}{c+1+ac}+\frac{1}{a+1+ab}\)

\(=\frac{ac+1}{c+1+ac}+\frac{1}{a+1+ab}\)

\(=\frac{ac+1}{c+abc+ac}+\frac{1}{a+1+ab}\)

\(=\frac{ac+1}{c\left(1+ab+a\right)}+\frac{1}{a+1+ab}\)

\(=\frac{ac+1}{c\left(1+ab+a\right)}+\frac{c}{c\left(a+1+ab\right)}\) \(MTC:c\left(a+1+ab\right)\)

\(=\frac{ac+1+c}{c\left(1+ab+a\right)}\)

\(=\frac{ac+abc+c}{c+abc+ac}\)

\(=1\)

Vậy \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\) là hằng số khi abc = 1 (đpcm)

Trịnh Trân Trân <3

a, A= (\(\dfrac{1}{3}\)-1)*(\(\dfrac{1}{6}\)-1)*(\(\dfrac{1}{10}\)-1)*(\(\dfrac{1}{15}\)-1)*(\(\dfrac{1}{21}\)-1)*(\(\dfrac{1}{28}\)-1)*(\(\dfrac{1}{36}\)-1)

b, B= (\(\dfrac{6}{8}\)+1)*(\(\dfrac{6}{18}\)+1)*(\(\dfrac{6}{30}\)+1)*...*(\(\dfrac{6}{10700}\)+1)

\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)

=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)

=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)

=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)

=\(\frac{1}{2014}\)

1.Tính :

a ) \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)

b ) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2017.2019}\)

c) \(S=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+....+\dfrac{1}{2018.2020}\)

d) \(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2017.2018.2019}\)

2. Tính tổng:

a) \(S=1.2+2.3+3.4+...+2018.2019\)

b) \(S=3.5+5.7+7.9+...+2017.2019\)

c) \(S=2.4+4.6+6.8+...+2018.2020\)

d) \(S=1.2.3+2.3.4+3.4.5+...+2017.2018.2019\)

3.Tính

a ) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2017.2020}\)

b ) \(S=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{2017.2019.2021}\)

Ai có công thức không cho mình xin với ????

1, biến đổi mỗi biểu thức sau thành một phân thức đại số
\(1\dfrac{1}{x}\) \(1+\dfrac{1}{1+\dfrac{1}{x}}\) \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}\) \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}\)

b, dự đoán kết quả của phép đổi biểu thức \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}}\) , \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}}}\) , \(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{x}}}}}}}}\) thành phân thức đại số và kiểm tra lại dự đoán đó
c. Xét quy luật của những phân thức đại số
d, cho vd minh họa