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) f(x) + g(x) = (x⁵ + 2x² - 1/2x² - 1/2x - 5) + (-x⁵ - 3x² + 1/2x + 1)

                     = x⁵ + 2x² - 1/2x² - 1/2x - 5 - x⁵ - 3x² + 1/2x + 1

                     = (x⁵ - x⁵) + (2x² - 1/2x² - 3x²) + (-1/2x + 1/2x) + (-5 + 1)

                     = -3/2x² - 4

f(x) - g(x) = (x⁵ + 2x² - 1/2x² - 1/2x - 5) - (-x⁵ - 3x² + 1/2x + 1)

                = x⁵ + 2x² - 1/2x² - 1/2x - 5 + x⁵ + 3x² - 1/2x - 1

                = (x⁵ + x⁵) + (2x² - 1/2x² + 3x²) + (-1/2 - 1/2x) + (-5 - 1)

                = 2x⁵ + 9/2x² - x - 6

b) f(x) + g(x) = -3/2x² - 4

Ta có:

-3/2x² > 0

=> -3/2x² - 4 > 1 > 0

=> f(x) + g(x) vô nghiệm

a, ta có:

\(f\left(x\right)=x^5+2x^2-\frac{1}{2}x^2-5\)

\(=x^5+\frac{3}{2}x^2-\frac{1}{2}x-5\)
\(f\left(x\right)+g\left(x\right)=-\frac{3}{2}x^2-4\)(t lm tắt nhé)

\(f\left(x\right)-g\left(x\right)=2x^5+\frac{9}{2}-x-6\)

b,Để  f(x)+g(x) có nghiệm thì

 \(f\left(x\right)+g\left(x\right)=-\frac{3}{2}x^2-4=0\)

\(\Rightarrow-\frac{3}{2}x^2=4\)

\(\Rightarrow x^2=-2\)(k tồn tại)

vậy f(x)+g(x) k có nghiệm.

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

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

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

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

a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)

\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)

\(=7x^4-9x^3+\frac{7}{4}x-3\)

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

\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)

\(=x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}\)

b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)

\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)

\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)

\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)

c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)

\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)

f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)

\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)

mi bn oi help me nha

a. f(x)+g(x)=2x⁵−4x⁴+3x³−x²+5x−1+

2x⁵−4x⁴+3x³−x²+5x−1+

−x⁵+2x⁴−3x³−x²−2x+7+x⁵−2x⁴−2x²−x−3

=-x⁵+x⁵+2x⁴-2x⁴-3x³-x²-2x²-2x-x+7-3

=-3x³-3x²-3x+4

d. f(x)-g(x)=2x⁵−4x⁴+3x³−x²+5x−1-(−x⁵+2x⁴−3x³−x²−2x+7)

=2x⁵−4x⁴+3x³−x²+5x−1-x⁵-2x⁴+3x³+x²+2x-7

=2x⁵-x⁵-4x⁴-2x⁴+3x³+3x³-x²+x²+5x+2x-1-7

=x⁵-6x⁴+6x³+7x-8

e. f(x)-h(x)=2x⁵−4x⁴+3x³−x²+5x−1-(x⁵−2x⁴−2x²−x−3)

=2x⁵−4x⁴+3x³−x²+5x−1-x⁵+2x⁴+2x²+x+3

=2x⁵-x⁵-4x⁴+2x⁴+3x³-x²+2x²+5x+x-1+3

=x⁵-2x⁴+3x³+x²+6x-4

h. g(x)-h(x)=−x⁵+2x⁴−3x³−x²−2x+7-(x⁵−2x⁴−2x²−x−3)

=−x⁵+2x⁴−3x³−x²−2x+7-x⁵+2x⁴+2x²+x+3

=-x⁵-x⁵+2x⁴+2x⁴-3x³-x²+2x²-2x+x+7+3

=-2x⁵+4x⁴-3x³+x²-x+10

f. f(x)+g(x)+h(x)=2x⁵−4x⁴+3x³−x²+5x−1+x⁵−2x⁴−2x²−x−3

=2x⁵-x⁵+x⁵-4x⁴+2x⁴-2x⁴+3x³-3x³-x²-x²-2x²+5x-2x-x-1+7-3

=2x⁵-4x⁴-4x²+2x+3

g. f(x)+g(x)-h(x)=2x⁵−4x⁴+3x³−x²+5x−1+x⁵−2x⁴−2x²−x−3)

=2x⁵−4x⁴+3x³−x²+5x−1+x⁵+2x⁴+2x²+x+3

=2x⁵-x⁵-x⁵-4x⁴+2x⁴+2x⁴+3x³-3x³-x²-x²+2x²+5x-2x+x-1+7+3

=4x+9

n. f(x)-g(x)+h(x)=2x⁵−4x⁴+3x³−x²+5x−1-x⁵−2x⁴−2x²−x−3

=2x⁵−4x⁴+3x³−x²+5x−1-x⁵−2x⁴−2x²−x−3

=2x⁵-x⁵+x⁵-4x⁴-2x⁴-2x⁴+3x³+3x³-x²+x²-2x²+5x+2x-x-1-7-3

=2x⁵-8x⁴+6x³-2x²+6x-11

m. f(x)-g(x)-h(x)=2x⁵−4x⁴+3x³−x²+5x−1-x⁵−2x⁴−2x²−x−3)

=2x⁵−4x⁴+3x³−x²+5x−1-x⁵+2x⁴+2x²+x+3

=2x⁵-x⁵-x⁵-4x⁴-2x⁴+2x⁴+3x³+3x³-x²+x²+2x²+5x+2x+x-1-7+3

=-4x⁴+6x³+2x²+8x-5

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