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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 \(a< b\Rightarrow b-a>0\Rightarrow A-B>0\Rightarrow A>B\)
Ta tính hiệu của M và T
ta có
Hiệu của Mẫu và Tử của A là 2019^2019-1 - (2019^2018-1) = 2019^2019 - 2019^2018 = 2019^2019.2018
Hiệu của Mẫu và Tử của B là 2019^2019+1 - (2019^2018+1) = 2019^2019 - 2019^2018 = 2019^2019.2018
2 Hiệu trên bằng nhau nên A < B
\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)
\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
Ta có :
\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)
\(\Rightarrow A-B< 0\)
\(\Rightarrow A< B\)
Vậy ta được \(A< B\)
a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)
Lời giải:
Ta có:
\(2018^{2018}(2019^{2019}+2019)=2018^{2018}.2019^{2019}+2018^{2018}.2019<2018^{2018}.2019^{2019}+2019^{2018}.2019 \)
\(< 2018^{2018}.2019^{2019}+2019^{2019}.2018\)
\(\Leftrightarrow 2018^{2018}(2019^{2019}+2019)< 2019^{2019}(2018^{2018}+2018)\)
\(\Rightarrow \frac{2018^{2018}}{2019^{2019}}< \frac{2018^{2018}+2018}{2019^{2019}+2019}\)
hây hây