Question

cho 2 đa thức

A(x)=\(x^5+2x^2-\frac{1}{2}x-3\)

B(X)=\(-x^5-3x^2+\frac{1}{2}x+1\)

a) tính M(x)=A(x)+B(x) ; N(x)=A(x)-B(x)

b) chứng tỏ M(x) ko có nghiệm

Answer 1

a)Ta có:M(x)=A(x)+B(x)

                  =\(\left(x^5+2x^2-\frac{1}{2}x-3\right)+\left(-x^5-3x^2+\frac{1}{2}x+1\right)\)

                  =\(x^5+2x^2-\frac{1}{2}x-3-x^5-3x^2+\frac{1}{2}x+1\)

                  =\(\left(x^5-x^5\right)+\left(2x^2-3x^2\right)+\left(-\frac{1}{2}x+\frac{1}{2}x\right)+\left(-3+1\right)\)

                  =\(-x^2-2\)

N(x)=A(x)-B(x)

      =\(x^5+2x^2-\frac{1}{2}x-3+x^5+3x^2-\frac{1}{2}x-1\)

      =\(2x^5+5x^2-x-4\)

b)M(x)=\(-x^2-2\)

Vì \(x^2\ge0\forall x\)

=>-\(x^2\)\(\le\)0\(\forall\)x

=>-\(x^2\)-2\(\le\)-2\(\forall\)x

=>-\(x^2\)-2<0

=>M(x)<0

vậy M(x) không có nghiệm