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\(x^4+2019x^2+2018x+2019\)
\(=x^4+x^2+1+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2018\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)
Câu a :
\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
Câu b :
\(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
\(x^2-3x+4\)
\(=x^2+x-4x+4\)
\(=\left(x^2+x\right)-\left(4x+4\right)\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x-4\right)\left(x+1\right)\).