Câu a:
\(x\left(3x+12\right)-7\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)
\(=20\)
Vậy ...
Câu b:
\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
\(=6x-3-5x+15+18x-24-19x\)
\(=-12\)
Vậy ...
a,\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x=20\)
b,\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x=6x-3-5x+15+18x-24-19x=-12\)