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

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

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

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

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

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

a, f (x) + g (x)

= \(-6x^2+x^3-8+12x+x^3-3x^2+6x-8\)

= \(-9x^2+2x^3-16+18x\)

g (x) - f (x)

= \(x^3-3x^2+6x-8+6x^2-x^3+8-12x\)

= \(3x^2-6x\)

b, Tính g (-1)

g (x) với x = (-1)

ta có : g (-1) = \((-1)^3-3.(-1)^2+6.(-1)-8\)

= \((-1)-3-6-8\)

= (-18)

Phần c thì sao ạ

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) 

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

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

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 ) 

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

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

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

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

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

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

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

Mọi người tk mình đi mình đang bị âm nè!!!!!!

Ai tk mình mình tk lại nha !!!

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 \(-\dfrac{4}{3}\)