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

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

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

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

\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^4+2x^2-\dfrac{1}{2}x\)

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

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

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

\(\Rightarrow g\left(x\right)=\dfrac{x^4+6x^2-5x}{2}\)

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

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

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

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

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

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

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

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

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

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

a, f(-1) =2.(-1)+1

=-2+1

=-1

g(-1)=1-2.-1

=1-(-2)

=3

2a,

f(0)=0² +7.0+12

=12

g(4)=4² -12

=4

1bta có :

f(g(x))=f(g(-1))

g(-1) =3=> f(3)

f(3)=2.3+1=7

g(f(x))=g(f(-1))

f(-1)=-1=> g(-1)

g(-1)=1-2.(-1)=3

Câu 2: 

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=2\\3^7\cdot a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{1093}\\b\simeq2\end{matrix}\right.\)

Câu 3: 

a: \(f\left(x\right)=4x^2+3x+1-3x^2+2x+3=x^2+5x+4\)

b: f(-4)=16-20+4=0

=>x=-4 là nghiệm 

c: Đặt f(x)=0

=>(x+4)(x+1)=0

=>x=-4 hoặc x=-1

f(x):g(x)

Để f(x) ⋮g(x) thì đa thức dư phải bằng 0 . do đó

\(\left\{{}\begin{matrix}a+21=0\\b+28=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-21\\b=-28\end{matrix}\right.\)

Vậy a=-21;b=-28 thì f(x) chia hết g(x)

\(a.\)

\(F\left(x\right)=2\left(x^4+x^3\right)+2x-4\left(x^2-x^3-1\right)+4\)

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

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

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

\(=5x^4-12-4x^4-2x^2+4x^3+2x^3-2x^2+2x\)

\(=x^4+6x^3-4x^2+2x-12\)

\(b.\)

\(K\left(x\right)=F\left(x\right)+G\left(x\right)\)

\(=\left(4x^4+6x^3-4x^2+2x-1\right)+\left(x^4+6x^3-4x^2+2x-12\right)\)

\(=5x^4+12x^3-8x^2+4x-13\)

\(H\left(x\right)=F\left(x\right)-G\left(x\right)\)

\(=\left(4x^4+6x^3-4x^2+2x-1\right)-\left(x^4+6x^3-4x^2+2x-12\right)\)

\(=4x^4+6x^3-4x^2+2x-1-x^4-6x^3+4x^2-2x+12\)

\(=3x^4+11\)

\(c.\)

Ta có : \(H\left(x\right)=36\)

\(\Rightarrow3x^4+11=36\)

\(\Rightarrow3x^4=25\)

1: f(-1)=0 

=>1+m-1+3m-2=0 và 

=>4m-2=0

=>m=1/2

2: g(2)=0

=>2^2-4(m+1)-5m+1=0

=>4-5m+1-4m-4=0

=>-9m+1=0

=>m=1/9

4: f(1)=g(2)

=>1-(m-1)+3m-2=4-4(m+1)-5m+1

=>1-m+1+3m-2=4-4m-4-5m+1

=>2m-2=-9m+1

=>11m=3

=>m=3/11

3:

H(-1)=0

=>-2-m-7m+3=0

=>-8m=-1

=>m=1/8

5: g(1)=h(-2)

=>1-2(m+1)-5m+1=-8-2m-7m+3

=>-5m+2-2m-2=-9m-5

=>-7m=-9m-5

=>2m=-5

=>m=-5/2