a)\(A=x^5-2018x^4+2018x^3-2018x^2+2018x-2019\)
\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2019\)
\(A=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2019\)
\(A=x-2019=2017-2019=-2\)
b)ta có:\(\left(x+1\right)^{20}+\left(y+2\right)^{30}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)
Thay vào \(\Rightarrow B=2\cdot\left(-1\right)^5+5\cdot\left(-2\right)^3+4\)
\(B=-2+\left(-40\right)+4=-38\)
thục hiền đc đó thục hiền ak nay vẫn hoc24 bình thường à 
Ta có x=2017 => 2018 = x+1 ; 2019= x+2
thay vào ta có : \(A=x^5-\left(x+1\right).x^4+\left(x+1\right).x^3-\left(x+1\right).x^2+\left(x+1\right).x-\left(x+2\right)\) \(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^x+x-x-2\) \(=\left(x^5-x^5\right)+\left(-x^4+x^4^{ }\right)+\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x\right)-2\)=-2
ey học tốt nhá
