a: \(A+B=x^3+\dfrac{3}{2}x+3\)
=>\(B=x^3+\dfrac{3}{2}x+3-x^4-2x^3+5x+10\)
=>\(B=-x^4-x^3+\dfrac{13}{2}x+13\)
b: \(A-C=2x^2\)
=>\(C=x^4+2x^3-5x-10-2x^2\)
c: D=A(2x-1)
\(=\left(x^4+2x^3-5x-10\right)\left(2x-1\right)\)
\(=2x^5-x^4+4x^4-2x^3-10x^2+5x-20x+10\)
\(=2x^5+3x^4-2x^3-10x^2-15x+10\)
d: A=(x+2)P
=>\(P=\dfrac{x^4+2x^3-5x-10}{x+2}=\dfrac{x^3\left(x+2\right)-5\left(x+2\right)}{x+2}=x^3-5\)
e: \(A=Q\cdot\left(x^3-5\right)\)
=>\(Q=\dfrac{x^4+2x^3-5x-10}{x^3-5}=\dfrac{x\left(x^3-5\right)+2\left(x^3-5\right)}{x^3-5}=x+2\)
=>Có đa thức Q thỏa mãn
e) \(A=x^4+2x^3-5x-10\)
\(A=x^3\left(x+2\right)-5\left(x+2\right)\)
\(A=\left(x+2\right)\left(x^3-5\right)\)
Vậy có đa thức \(Q\left(x\right)=x+2\) để \(A=\left(x^3-5\right)Q\)
f) Theo trên ta được \(P\left(x\right)=x^3-5\) để \(A=\left(x+2\right).P\)
