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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 : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
ta có:\(\left|x-1000\right|+\left|x-2019\right|=\left|-x+1000\right|+\left|x-2019\right|\)
\(\ge\left|-x+1000+x-2019\right|=1019\)
dấu = xảy ra khi \(\left(-x+1000\right).\left(x-2019\right)\ge0\)
\(\Rightarrow1000\le x\le2019\)
\(\hept{\begin{cases}\left|x-2018\right|\ge0\\\left|y-10\right|\ge0\\\left|z-1\right|\ge0\end{cases}}\text{dấu = xảy ra khi }\hept{\begin{cases}x=2018\\y=10\\z=1\end{cases}}\)
Vậy để \(\left|x-1000\right|+\left|x-2018\right|+\left|x-2019\right|+\left|y-10\right|+\left|z-1\right|=1019\) => \(\hept{\begin{cases}x=2018\\y=10\\z=1\end{cases}}\)
\(=3\cdot2+\left(-999\right)^{10}-1=999^{10}+17\)