Bài 1:
a) Ta có: \(f\left(x\right)=x^2+2x^4+10x-3x^2+x^2-x+5\)
\(=2x^4-x^2+9x+5\)
Ta có: \(g\left(x\right)=x-5x-x^2-x^4+3x+x^2-2x^2-2x^3-3x\)
\(=-x^4-2x^3-2x^2-4x\)
b) Ta có: \(f\left(x\right)+g\left(x\right)\)
\(=2x^4-x^2+9x+5-x^4-2x^3-2x^2-4x\)
\(=x^4-2x^3-3x^2+5x+5\)
Ta có: \(f\left(x\right)-g\left(x\right)\)
\(=2x^4-x^2+9x+5+x^4+2x^3+2x^2+4x\)
\(=3x^4+2x^3+x^2+13x+5\)
c) Ta có: \(f\left(x\right)+g\left(x\right)=x^4-2x^3-3x^2+5x+5\)
nên khi x=-1 thì \(f\left(-1\right)+g\left(-1\right)=\left(-1\right)^4-2\cdot\left(-1\right)^3-3\cdot\left(-1\right)^2+5\cdot\left(-1\right)+5\)
\(=1+2-3-5+5\)
\(=0\)
Ta có: \(f\left(x\right)-g\left(x\right)=3x^4+2x^3+x^2+13x+5\)
nên khi x=-1 thì \(f\left(-1\right)-g\left(-1\right)=3\cdot\left(-1\right)^4+2\cdot\left(-1\right)^3+\left(-1\right)^2+12\cdot\left(-1\right)+5\)
\(=3+2+1-12+5\)
\(=-1\)