\(a)\)
\(f\left(x\right)=-x^3+3x^2+x-3+2^3-x^2\)
\(\Leftrightarrow f\left(x\right)=-x^3+\left(3x^2-x^2\right)+x-3+2^3\)
\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x-3+8\)
\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x+5\)
\(g\left(x\right)=-3x^3-x^2+2x^3+5x-3-4x\)
\(\Leftrightarrow g\left(x\right)=\left(-3x^3+2x^3\right)-x^2+\left(5x-4x\right)-3\)
\(\Leftrightarrow g\left(x\right)=-x^3-x^2+x-3\)
\(b)\)
Theo đề ra: \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)
\(f\left(x\right)=-x^3+2x^2+x+5\)
\(g\left(x\right)=-x^3-x^2+x-3\)
\(\Rightarrow h\left(x\right)=x^2+2x+2\)
\(c)\)
\(f\left(2\right)=-2^3+2.2^2+2+5\)
\(\Leftrightarrow f\left(2\right)=-8+4^2+7\)
\(\Leftrightarrow f\left(2\right)=-8+16+7\)
\(\Leftrightarrow f\left(2\right)=15\)
\(f\left(-1\right)=-\left(-1\right)^3+2\left(-1\right)^2+\left(-1\right)+5\)
\(\Leftrightarrow f\left(-1\right)=1+-2^2-1+5\)
\(\Leftrightarrow f\left(-1\right)=1+4+4\)
\(\Leftrightarrow f\left(-1\right)=9\)
\(d)\)
\(x^2+2x+2=0\)
\(\Leftrightarrow x^2+2\left(x+1\right)=0\)
Ta có: \(x^2>0\forall x\)
\(x^2+2\left(x+1=0\right)\)
\(\Leftrightarrow x^2=-2\left(x+1\right)\)(Loại)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
\(\Leftrightarrow2\left(x+1\right)=0\Leftrightarrow x=\left(-1\right)\)