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Có \(a\left(b+1\right)< b\left(a+1\right)\Leftrightarrow ab+a< ab+b\)

\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)

Áp dụng \(\frac{2^{2018}}{3^{2019}}< \frac{2^{2018}+1}{3^{2019}+1}\)

Ta có:

\(1-\frac{a}{b}=\frac{b-a}{b}\)

\(1-\frac{a+1}{b+1}=\frac{b+1-a-1}{b+1}=\frac{b-a}{b+1}\)

Vì b < b + 1 và a < b; a, b nguyên dương  => b - a > 0 nên \(\frac{b-a}{b}>\frac{b-a}{b+1}\)

Do đó \(1-\frac{a}{b}>1-\frac{a+1}{b+1}\)

\(\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\)

Áp dụng chứng minh tương tự nhé bạn

Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)

Áp dụng :

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)

\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)

Sr còn thiếu

\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)

Vậy \(A< B\)

Ta có:

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)

\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)

\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)

\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)

\(A=\frac{1.2018}{2017.2}\)

\(A=\frac{1009}{2017}\)

Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)

           \(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)

Vậy A>B

Lời giải:

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{100}}$

$\Rightarrow 2A=1-\frac{1}{3^{100}}<1$

$\Rightarrow A< \frac{1}{2}$

$\Rightarrow A< B$

\(\frac{a^2}{1+a+a^2}\)

\(\frac{1}{1+a}\)

\(\frac{b^2}{1+b+b^2}\)=\(\frac{1}{1+b}\)

vì a>b nên  \(\frac{a^2}{1+a+a^2}\)>\(\frac{b^2}{1+b+b^2}\)

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)

\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)

\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)

\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)

\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)

Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)

a) Ta có: a < b => a + 1 < b + 1

b) Ta có: a < b => a - 2 < b - 2