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\(\left(a.b\right)^n=\left(a.b\right)\left(a.b\right)\left(a.b\right)...\left(a.b\right)\) (n thừa số ) \(=\left(a.a.a...a\right)\left(b.b...b\right)=a^n.b^n\)
[ a.b ] = [a.b ] [ a.b ] [ a.b ] ... [ a.b ] n là thừa số = [ a.a.a...a ] [ b.b...b ] = a n.b n
a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)
b,
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)
\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)
\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)
\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)
a. Ta có
\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}.\)
Vì\(\frac{2011}{2012+2013}< \frac{2011}{2012}.\)(1)
\(\frac{2012}{2012+2013}< \frac{2012}{2013}.\)(2)
Cộng vế với vế của 1;2 ta được
\(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}< A=\frac{2011}{2012}+\frac{2012}{2013}\)
hay A>B
Tham khảo:
Câu hỏi của nguyễn thùy linh - Toán lớp 6 - Học toán với OnlineMath
nhé!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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\(B-A=\frac{11-10}{a^m}+\frac{9-10}{a^n}=\frac{1}{a^m}-\frac{1}{a^n}\)
Nếu \(m>n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}< 0\Rightarrow B< A\)
Nếu \(m< n\) thì \(\frac{1}{a^m}-\frac{1}{a^n}>0\Rightarrow B>A\)
a) Ta có (am)n = am.am...am (định nghĩa) (có n thừa số am)
= am + m + .... + m (có n hạng tử m)
= am.n (đpcm)
b) Ta có 5333 = 53.111 = (53)111 = 125111
3555 = 35.111 = (35)111 = 243111
Nhận thấy 125 < 243
=> 125111 < 243111
=> 5333 < 3555
b) Ta có 2400 = 24.100 = (24)100 = 16100
4200 = 42.100 = (42)100 = 16100
=> 2400 = 4200 (= 16100)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(B=\frac{510}{103}\)