Question

[Nguyễn Hải Đăng]

Cho A=\(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}\) và B=\(\dfrac{a^{2019}-b^{2019}}{a^{2019}+b^{2019}}\)

So sánh A với B.

Answer 1

[Nguyễn Việt Lâm]

Giải trâu:

Xét \(A-B=\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}-\dfrac{a^{2019}-b^{2019}}{a^{2019}+b^{2019}}\)

\(=\dfrac{\left(a^{2018}-b^{2018}\right)\left(a^{2019}+b^{2019}\right)-\left(a^{2018}+b^{2018}\right)\left(a^{2019}-b^{2019}\right)}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)

\(=\dfrac{a^{4037}+a^{2018}b^{2019}-a^{2019}b^{2018}-b^{4037}-a^{4037}+a^{2018}b^{2019}-a^{2019}b^{2018}+b^{4037}}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)

\(=\dfrac{2a^{2018}b^{2019}-2a^{2019}b^{2018}}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}=\dfrac{2a^{2018}b^{2018}\left(b-a\right)}{\left(a^{2018}+b^{2018}\right)\left(a^{2019}+b^{2019}\right)}\)

\(\Rightarrow\)Nếu \(a>b\Rightarrow b-a< 0\Rightarrow A-B< 0\Rightarrow A< B\)

Nếu \(a< b\Rightarrow b-a>0\Rightarrow A-B>0\Rightarrow A>B\)