=> 2 f(x) = 6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9

               = 10x^4 - 6x^3 + 4x^2 + 8x - 14 

=> 2.f ( x ) =  2 ( 5x^4 - 3x^3 + 2x^2 + 4x - 7 )

=> ( fx) = 5x^4 - 3x^3 + 2x^2 + 4x - 7 

g(x) tự tìm 

ta có:

f(x) + g(x) = 6x^4 - 3x^2 - 5 

f(x) - g(x) = 4x^4 - 6x^3 + 7x^2 + 8x - 9

công hai vế lại với nhau ta được:

f(x)+g(x)+f(x)-g(x)=6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9

=>2f(x)=6x⁴+4x⁴-6x³-3x²+7x²+8x-5-9

2f(x)=10x⁴-6x³+4x²+8x-14

2f(x)=2.(5x⁴-3x³+2x²+4x-7)

=>f(x)=5x⁴-3x³+2x²+4x-7

=>g(x)=6x^4 - 3x^2 - 5 -(5x⁴-3x³+2x²+4x-7)

=6x⁴-3x²-5-5x⁴+3x³-2x²-4x+7

=6x⁴-5x⁴+3x³-3x²-2x²-4x-5+7

=x⁴+3x³-5x²-4x+2

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

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

\(f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\)

suy ra:  \(\left[f\left(x\right)+g\left(x\right)\right]+\left[f\left(x\right)-g\left(x\right)\right]=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\)

\(\Leftrightarrow\)\(2f\left(x\right)=10x^4-6x^3+4x^2+8x-14\)

\(\Rightarrow\)\(f\left(x\right)=5x^4-3x^3+2x^2+4x-7\)

       \(g\left(x\right)=6x^4-3x^2-5-f\left(x\right)\)

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

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\\ \Rightarrow2f\left(x\right)=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\\ 2f\left(x\right)=10x^4-6x^3+4x^2+8x-14\\ 2f\left(x\right)=2\left(5x^4-3x^3+2x^2+4x-7\right)\\ \Rightarrow f\left(x\right)=5x^4-3x^3+2x^2+8x-14\)

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ \Rightarrow g\left(x\right)=6x^4-3x^2-5-f\left(x\right)\\ g\left(x\right)=6x^4-3x^2-5-5x^4+3x^3-2x^2-8x+14\\ g\left(x\right)=x^4+3x^3-5x^2-8x+9\)