a, f(x)+g(x)= (\(x^5-3\) + 7\(x^4-9x^3+x^2-\dfrac{1}{4}x\))+(\(5x^4-x^5\)+\(x^2\)\(-2x^3+3x^2-\dfrac{1}{4})\)

= \(12x^4-12x^3+5x^2-\dfrac{1}{4}x-\dfrac{13}{4}\)

b, f(x)\(-\)g(x)= (\(x^5-3\) + 7\(x^4-9x^3+x^2-\dfrac{1}{4}x\))\(-\)(\(5x^4-x^5\)+\(x^2\)\(-2x^3+3x^2-\dfrac{1}{4})\)

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

=2x\(^5\)+2x\(^4\)\(-7x^3\)\(-2x^2\)\(-\dfrac{1}{4}x\) \(-\) \(\dfrac{11}{4}\)

c,Ta có:h(x)+f(x)=f(x) \(\Rightarrow\)h(x)=f(x)\(-\)f(x)=0

a, 4x^3 +3x^2+7x

b, = 0

h(x)=f(x)+g(x)=(1+1)x^3 +(2-2)x^2+(-15+5)

h(x)=2x^3-10

h(x) + g(x) = f(x)

=> h(x)= f(x) - g(x) = \(3x^4+2x^2-2x^4+x^2-5x-\left(x^4-x^2-2x+6+3x^2\right)=x^2-3x-6\)\(h\left(-\dfrac{1}{3}\right)=\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)-6=\dfrac{-44}{9}\)

\(h\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^2-3\cdot\dfrac{3}{2}-6=-\dfrac{33}{4}\)

\(x^2-3x-6=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{6}\\x=\dfrac{3-\sqrt{33}}{6}\end{matrix}\right.\)

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:\(h\left(x\right)=3x^4+5x^3-x^2+2x+17.\)

Giải:

a) \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

\(\Leftrightarrow h\left(x\right)=9-x^5+4x-2x^3+x^2-7x^4+x^5-9+2x^2+7x^4+2x^3-3x\)

\(\Leftrightarrow h\left(x\right)=x+3x^2\)

b) Để đa thức h(x) có nghiệm

\(\Leftrightarrow h\left(x\right)=0\)

\(\Leftrightarrow x+3x^2=0\)

\(\Leftrightarrow x\left(1+3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

a)

         f(x)= -x⁵ -7x⁴ -2x³+ x² + 4x + 8

         g(x)=x⁵ +7x⁴+2x³+3x² - 3x   -8

f(x)+g(x)  =0   -0    -0    + 4x² +x+0

         g(x)=x⁵ +7x⁴+2x³+3x² - 3x  -8

         f(x)= -x⁵ -7x⁴ -2x³+ x² + 4x + 8

g(x)-f(x)  =2x⁵+14x⁴+4x³+2x²-7x  -16

b)

Bậc:5

Hệ số cao nhất:2

hệ số tự do:16

c)

Để đt h(x) có nghiệm thì 

4x²+x=0

->4x.x+x=0

->(4x+1)x=0

->th1:x=0 -> x=0

        4x+1=0 -> x=-1/4

Vậy đt h(x) có nghiệm là x=0 hoặc x=-1/4

Lần sau bn viết rõ hơn nhé

mik dich mún lòi mắt

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