Giải:
a) Thu gọn và sắp xếp:
\(F\left(x\right)=5x^2-1+3x+x^2-5x^3\)
\(\Leftrightarrow F\left(x\right)=6x^2-1+3x-5x^3\)
\(\Leftrightarrow F\left(x\right)=-5x^3+6x^2+3x-1\)
\(G\left(x\right)=2-3x^3+6x^2+5x-2x^3-x\)
\(\Leftrightarrow G\left(x\right)=2-5x^3+6x^2+4x\)
\(\Leftrightarrow G\left(x\right)=-5x^3+6x^2+4x+2\)
b) \(M\left(x\right)=F\left(x\right)-G\left(x\right)\)
\(\Leftrightarrow M\left(x\right)=-5x^3+6x^2+3x-1-\left(-5x^3+6x^2+4x+2\right)\)
\(\Leftrightarrow M\left(x\right)=-5x^3+6x^2+3x-1+5x^3-6x^2-4x-2\)
\(\Leftrightarrow M\left(x\right)=-x-3\)
\(N\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(\Leftrightarrow N\left(x\right)=-5x^3+6x^2+3x-1+\left(-5x^3+6x^2+4x+2\right)\)
\(\Leftrightarrow N\left(x\right)=-5x^3+6x^2+3x-1-5x^3+6x^2+4x+2\)
\(\Leftrightarrow N\left(x\right)=-10x^3+12x^2+7x+1\)
c) Để đa thức M(x) có nghiệm
\(\Leftrightarrow M\left(x\right)=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
\(\Leftrightarrow x=-3\)
Vậy ...
