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a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
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 \(a< b\Rightarrow b-a>0\Rightarrow A-B>0\Rightarrow A>B\)
Lời giải:
Ta có:
\(2018^{2018}(2019^{2019}+2019)=2018^{2018}.2019^{2019}+2018^{2018}.2019<2018^{2018}.2019^{2019}+2019^{2018}.2019 \)
\(< 2018^{2018}.2019^{2019}+2019^{2019}.2018\)
\(\Leftrightarrow 2018^{2018}(2019^{2019}+2019)< 2019^{2019}(2018^{2018}+2018)\)
\(\Rightarrow \frac{2018^{2018}}{2019^{2019}}< \frac{2018^{2018}+2018}{2019^{2019}+2019}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Rightarrow a=b=c=1\)
\(Ta\) \(có\) :
\(\dfrac{1^3\cdot1^2\cdot1^{2018}}{1^{2019}}=1\)
\(=3\cdot2+\left(-999\right)^{10}-1=999^{10}+17\)