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 

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

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

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

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

Bài 1:

a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)

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

\(=2x-5\)

Bài 1: 

b) 

\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)

\(P\left(3\right)=2\cdot3-5=6-5=1\)

Bài 2: 

a) Ta có: f(x)+g(x)

\(=x^2-6x+4+x^2-4x-2\)

\(=2x^2-10x+2\)

a) A(x) = f(x) + g(x)

= (3x⁴ - 5 + 2x⁵ - 6x³ + 2x² + 4x) + (3x - x² + 5 - 2x⁵ - 3x⁴ + 6x³)

= 3x⁴ - 5 + 2x⁵ - 6x³ + 2x² + 4x + 3x - x² + 5 - 2x⁵ - 3x⁴ + 6x³

= x² + 7x

Vậy A(x) = x² + 7x

b) Đặt A(x) = 0, ta có:

A(x) = x² + 7x = 0

=> x(x + 7) = 0

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+7=0\Rightarrow x=-7\end{matrix}\right.\)

Vậy nghiệm của A(x) là x = 0 hoặc x = -7

Áp dụng quy tắc tổng hiệu đó

\(f\left(x\right)=\dfrac{\left(x^3+6x^2+3x^4\right)+\left(2x^3-x^2+3x^4\right)}{2}\)

Vậy \(f\left(x\right)=\dfrac{6x^4+3x^3+5x^2}{2}=3x^4+1,5x^3+2,5x^2\)

\(g\left(x\right)=\left(x^3+6x^2+3x^4\right)-f\left(x\right)\)

\(=\left(x^3+6x^2+3x^4\right)-\left(3x^4+1,5x^3+2,5x^2\right)\)

\(=x^3+6x^2+3x^4-3x^4-1,5x^3-2,5x^2\)

\(=\left(3x^4-3x^4\right)+\left(x^3-1,5x^3\right)+\left(6x^2-2,5x^2\right)\)

Vậy \(g\left(x\right)=-0,5x^3+3,5x^2\)

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

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

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

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

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

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

a) f(x) + g(x) = \(5x^2-2x+5+5x^2-6x-\dfrac{1}{3}=10x^2-8x+\dfrac{14}{3}\)

b) f(x) - g(x) = \(5x^2-2x+5-5x^2+6x+\dfrac{1}{3}=4x+\dfrac{16}{3}\)

c) Ngiệm của f(x) - g(x) chính là nghiệm của \(4x+\dfrac{16}{3}\)

Ta có: \(4x+\dfrac{16}{3}=0\Leftrightarrow4x=-\dfrac{16}{3}\Leftrightarrow x=-\dfrac{4}{3}\)

Vậy nghiệm của f(x) - g(x) là \(-\dfrac{4}{3}\)