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

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

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

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

Ta có công thức: \(1+2+3+4+...+n=\frac{n\cdot\left(n+1\right)}{2}\)

Ta có:\(\frac{1+2+3+...+a}{a}< \frac{1+2+3+...+b}{b}\)

\(\Leftrightarrow\frac{\frac{a\left(a+1\right)}{2}}{a}< \frac{\frac{b\left(b+1\right)}{2}}{b}\)

\(\Leftrightarrow\frac{a\left(a+1\right)}{2a}< \frac{b\left(b+1\right)}{2b}\)

\(\Leftrightarrow\frac{a+1}{2}< \frac{b+1}{2}\)

\(\Leftrightarrow a+1< b+1\)

\(\Leftrightarrow a< b\)

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 \(A< B\)

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

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

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

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

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

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

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

ta co:

2A=2(2 mu 60 +1 /2 mu 61 +1)

2A=2 mu 61 +2 / 2 mu 61 +1

2A=2 mu 61 +1+1/2 mu 61 +1

2A=1+1/2 mu 61 +1

ta co:

2B=2(2 mu 61 +1/2 mu 62 +1)

2B=2 mu 62 +2/2 mu 62+1

2B=2 mu 62 +1+1/2 mu 62 +1

2B=1+1/2 mu 62 +1

mà 1+1/2 mu 61+1>1+1/2 mu 62 +1 nen 2A >2B

vậy A>B

nhớ k đúng cho mk nha

Ta có:

2.A=2 mủ 61 +2/2 mủ 61 +1=1+(2/2 mủ 61 +1)

2.B=2 mủ 62 + 2 /2 mủ 62 +1=1+(2/2 mủ 62 + 1)

vì ... nên 2.A >2.B.Vậy A>B

Bài 1 :

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

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

\(=1-\frac{1}{50}< 1\left(1\right)\)

\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)

Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)

Bài 2:

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)

\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)

\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)

\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)

Chúc bạn học tốt ( -_- )

Bài 1:

ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}< 1\)

\(\Rightarrow A< 1\)(1) 

ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)

                                                                               \(=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)

\(\Rightarrow B>1\)(2)

Từ (1);(2) => A<B

Bài 2:

ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(\Rightarrow S=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(S=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

thay số: \(S=7.\frac{7}{10}-3\)

\(S=4\frac{9}{10}-3\)

\(S=1\frac{9}{10}=\frac{19}{10}\)

mà  \(1\frac{8}{11}=\frac{19}{11}\)

\(\Rightarrow\frac{19}{10}>\frac{19}{11}\)

\(\Rightarrow S>\frac{19}{11}\)

\(\Rightarrow S>1\frac{8}{11}\)

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)

Ta có :

\(\frac{1}{2^2}-1>-\frac{1}{2}\)

\(\frac{1}{3^2}-1>-\frac{1}{2}\)

...........

\(\frac{1}{2018^2}-1>\frac{1}{2}\)

\(\Rightarrow A>B\)