1) \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
\(=\left(x+3\right).x^2-5\left(x+3\right)+\left(x+4\right)\left(x-1x^2\right)\)
\(=x^3+3x^2-5x-15+\left(x+4\right)\left(x-x^2\right)\)
\(=x^3+3x^2-5x-15-x^3+x^2-4x^2+4x\)
\(=3x^2-5x-15-3x^2+4x\)
\(=-x-15\)
2) Đặt đa thức là \(N\left(x\right)\)ta được: \(3x^3+2x^2-x+k=N\left(x\right)\left(x-1\right)\)
Để \(3x^3+2x^2-x+K⋮x-1\Leftrightarrow x=1\)
Thay vào ta được
\(\Rightarrow3.1^3+2.1^2-1+K=0\)
\(\Rightarrow3+2-1+K=0\)
\(\Rightarrow K=-4\)
3) \(Q=x^2-x+3\)
\(=x^2-x+\frac{1}{4}+\frac{11}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)
Với \(\forall x\) ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow N=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)