b. Ta có f(x) + 2g(x)

= x3 - 2x2 + 2x- 5 + 2(-x3 + 3x2 - 2x + 4)

= x3 - 2x2 + 2x - 5 + (-2x3) + 6x2 - 4x + 8

=-x3 + 4x2 - 2x + 3 (0.5 điểm)

2f(x) - g(x) = x3 - 2x2 + 2x- 5 - 2(-x3+ 3x2 - 2x + 4)

= x3 - 2x2 + 2x - 5 + 2x3 - 6x2 + 4x - 8

= 3x3 - 8x2 + 6x - 13 (0.5 điểm)

Ta có:\(f\left(x\right)-h\left(x\right)=g\left(x\right)\Leftrightarrow h\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(\Leftrightarrow h\left(x\right)=\left(2x^4+5x^3-x+8\right)-\left(x^4-x^2+3x+9\right)\)

                  \(=2x^4+5x^3-x+8-x^4-x^2-3x-9\)

                  \(=x^4+5x^3+x^2-4x-1.\)

Vậy, đa thức cần tìm là: \(h\left(x\right)=x^4+5x^3+x^2-4x-1.\)

Ta có:  \(h\left(x\right)-g\left(x\right)=f\left(x\right)\Leftrightarrow h\left(x\right)=f\left(x\right)+g\left(x\right)\)

\(\Leftrightarrow h\left(x\right)=\left(2x^4+5x^3-x+8\right)+\left(x^4-x^2+3x+9\right)\)

                  \(=2x^4+5x^3-x+8+x^4-x^2+3x+9\)

                  \(=3x^4+5x^3-x^2+2x+17\)

Vậy, đa thức cần tìm là:\(h\left(x\right)=3x^4+5x^3-x^2+2x+17.\)

1:

a: f(x)=2x^4+2x^3+2x^2+5x+6

g(x)=x^4-2x^3-x^2-5x+3

c: h(x)=2x^4+2x^3+2x^2+5x+6+x^4-2x^3-x^2-5x+3=3x^4+x^2+9

K(x)=f(x)-2g(x)-4x^2

=2x^4+2x^3+2x^2+5x+6-2x^4+4x^3+2x^2+10x-6-4x^2

=6x^3+15x

c: K(x)=0

=>6x^3+15x=0

=>3x(2x^2+5)=0

=>x=0

d: H(x)=3x^4+x^2+9>=9

Dấu = xảy ra khi x=0

a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)

=\(-x^5+2x^4-4x^2-1\)

f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

=\(3x^5-10x^4-13\)

b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)

=\(x^4+9x^3-11x^2+7x-2\)

f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)

=\(9x^4+5x^3-x^2-x-12\)

a ) 

\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)

\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)

b ) Làm tương tự 

a) Ta có: \(f\left(x\right)=5x^4+x^3-x+11+x^4-5x^3\)

\(=\left(5x^4+x^4\right)+\left(x^3-5x^3\right)-x+11\)

\(=6x^4-4x^3-x+11\)

Ta có: \(g\left(x\right)=2x^2+3x^4+9-4x^2-4x^3+2x^4-x\)

\(=\left(3x^4+2x^4\right)-4x^3+\left(2x^2-4x^2\right)-x+9\)

\(=5x^4-4x^3-2x^2-x+9\)

b) Ta có: h(x)=f(x)-g(x)
\(=6x^4-4x^3-x+11-5x^4+4x^3+2x^2+x-9\)

\(=x^4+2x^2+2\)

c) Ta có: \(x^4\ge0\forall x\)

\(2x^2\ge0\forall x\)

Do đó: \(x^4+2x^2\ge0\forall x\)

\(\Leftrightarrow x^4+2x^2+2\ge2>0\forall x\)

Vậy: Đa thức h(x) không có nghiệm(Đpcm)

Xét [\(f\left(x\right)+g\left(x\right)\)]+[\(f\left(x\right)-g\left(x\right)\)]=\(\left[2x^4+5x^2-3x\right]\)+\(\left[x^4-x^2+2x\right]\)

\(2f\left(x\right)=2x^4+5x^2-3x+x^4-x^2+2x\)

\(2f\left(x\right)=3x^4+4x^2-x\)

\(\Rightarrow f\left(x\right)=\dfrac{3x^4+4x^2-x}{2}\)

\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^4+2x^2-\dfrac{1}{2}x\)

Xét \(\left[f\left(x\right)+g\left(x\right)\right]-\left[f\left(x\right)-g\left(x\right)\right]=\)\(\left[2x^4+5x^2-3x\right]\)\(-\)\(\left[x^4-x^2+2x\right]\)

\(2g\left(x\right)=\)\(2x^4+5x^2-3x-x^4+x^2-2x\)

\(2g\left(x\right)=x^4+6x^2-5x\)

\(\Rightarrow g\left(x\right)=\dfrac{x^4+6x^2-5x}{2}\)

\(\Rightarrow g\left(x\right)=\dfrac{1}{2}x^4+3x^2-\dfrac{5}{2}x\)