3/a)\(M\left(x\right)=\frac{1}{2}x^2-3x-x^3+3\)
\(=-x^3+\frac{1}{2}x^2-3x+3\)
\(N\left(x\right)=-4x+x^2+\frac{1}{2}x^3+6\)
\(=\frac{1}{2}x^3+x^2-4x^3+6\)
b)Ta có:\(\text{A}\left(x\right)=M\left(x\right)-N\left(x\right)\)
hay \(\text{A}\left(x\right)=\left(-x^3+\frac{1}{2}x^2-3x+3\right)-\left(\frac{1}{2}x^3+x^2-4x+6\right)\)
\(=-x^3+\frac{1}{2}x^2-3x+3-\frac{1}{2}x^3-x^2+4x-6\)
\(=-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3\)
Đặt\(\text{A}\left(x\right)=0\)
\(\Rightarrow-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3=0\)
\(-\frac{3}{2}x^3-\frac{1}{2}x^2=-x+3\)
\(-2\left(x^3-x^2\right)=-x+3\)
\(x^3-x^2+x=3+2=5\)
\(x^2=5\)
\(\Rightarrow x=\sqrt{5}\)
Vậy \(\text{A}\left(x\right)\) có 1 nghiệm là \(\sqrt{5}\)
