\(\text{a)}P\left(x\right)=2x^2+2x-6x^2+4x^3+2-x^3\)

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

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

\(Q\left(x\right)=3x^3+2x+3\)

\(\text{b)}C\left(x\right)=P\left(x\right)+Q\left(x\right)\)

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

                 \(Q\left(x\right)=3x^3\)                \(2x+3\)

\(P\left(x\right)+Q\left(x\right)=6x^3-4x^2+4x+5\)

             \(\Rightarrow C\left(x\right)=6x^3-4x^2+4x+5\)

\(\text{c)}D\left(x\right)=Q\left(x\right)-P\left(x\right)\)

                 \(Q\left(x\right)=3x^3\)                \(2x+3\)

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

\(Q\left(x\right)-P\left(x\right)=\)       \(4x^2\)             \(+1\)

             \(\Rightarrow D\left(x\right)=4x^2+1\)

Để \(D\left(x\right)\)có nghiệm thì:

         \(D\left(x\right)=0\)

\(\Rightarrow4x^2+1=0\)

Mà \(4x^2\ge0\)

\(\Rightarrow4x^2+1\ge1\)

\(\Rightarrow D\left(x\right)\ge1\)

\(\Rightarrow D\left(x\right)>0\)

Vậy đa thức \(D\left(x\right)\)vô nghiệm

a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)

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

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

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

b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)

\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)

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

\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)

\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`P(x) =`\(3x^2+7+2x^4-3x^2-4-5x+2x^3\)

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

`= 2x^4 + 2x^3 - 5x + 3`

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

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

`= 4x^4 + 4x^3 + 2x^2 + 5x - 2`

`b)`

`P(-1) = 2*(-1)^4 + 2*(-1)^3 - 5*(-1) + 3`

`= 2*1 + 2*(-1) + 5 + 3`

`= 2 - 2 + 5 + 3`

`= 8`

___

`Q(0) = 4*0^4 + 4*0^3 + 2*0^2 + 5*0 - 2`

`= 4*0 + 4*0 + 2*0 + 5*0 - 2`

`= -2`

`c)`

`G(x) = P(x) + Q(x)`

`=> G(x) = 2x^4 + 2x^3 - 5x + 3 + 4x^4 + 4x^3 + 2x^2 + 5x - 2`

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

`= 6x^4 + 6x^3 + 2x^2 + 1`

`d)`

`G(x) = 6x^4 + 6x^3 + 2x^2 + 1`

Vì `x^4 \ge 0 AA x`

    `x^2 \ge 0 AA x`

`=> 6x^4 + 2x^2 \ge 0 AA x`

`=> 6x^4 + 6x^3 + 2x^2 + 1 \ge 0`

`=> G(x)` luôn dương `AA` `x`

`a,`

`P(x)=2x^3-2x+x^2-x^3+3x+2`

`= (2x^3-x^3)+x^2+(-2x+3x)+2`

`= x^3+x^2+x+2`

`b,`

`H(x)+Q(x)=P(x)`

`-> H(x)=P(x)-Q(x)`

`-> H(x)=(x^3+x^2+x+2)-(x^3-x^2-x+1)`

`H(x)=x^3+x^2+x+2-x^3+x^2+x-1`

`= (x^3-x^3)+(x^2+x^2)+(x+x)+(2-1)`

`= 2x^2+2x+1`

Vậy, `H(x)=2x^2+2x+1.`

a.

\(P\left(x\right)=x^3+x^2+x+2\)

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

b.

\(H\left(x\right)+Q\left(x\right)=P\left(x\right)\Rightarrow H\left(x\right)=P\left(x\right)-Q\left(x\right)\)

\(\Rightarrow H\left(x\right)=x^3+x^2+x+2-\left(x^3-x^2-x+1\right)\)

\(\Rightarrow H\left(x\right)=2x^2+2x+1\)

\(a,Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\\ =\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2+\left(3x-3x\right)+1\\ =3x^4+2x^2+1\\ b,Q\left(x\right)=0\\ \Leftrightarrow3x^4+2x^2+1=0\\ \Delta=b^2-4ac=2^2-4.3.1=-8< 0\)

Vậy Q(x) không có nghiệm

a) P(x) = -2x^2 + 4x^4 – 9x^3 + 3x^2 – 5x + 3

=4x^4-9x^3+x^2-5x+3

Q(x) = 5x^4 – x^3 + x^2 – 2x^3 + 3x^2 – 2 – 5x

=5x^4-3x^3+4x^2-5x-2

b)

P(x)

-bậc:4

-hệ số tự do:3

-hệ số cao nhất:4

Q(x)

-bậc :4

-hệ số tự do :-2

-hệ số cao nhất:5

a, P(x)=(2x^3-x^3)+x^2+(3x-2x)+2=x^3+x^2+x+2
Q(x)=(3x^3-4x^3)+(5x^2-4x^2)+(3x-4x)+1=-x^3+x^2-x+1
b, M(x)=P(x)+Q(x)=x^3+x^2+x+2+(-x^3)+x^2-x+1=2x^2+3
N(x)=P(x)-Q(x)=x^3+x^2+x+2-(-x^3+x^2-x+1)=2x^3+2x+1
c, M(x)=2x^2+3
do x^2>=0 với mọi x=2x^2>=0
nên 2x^2+3>=3 với mọi x
để M(x) có nghiệm thì phải tồn tại x để M(x)=0 ( vô lý vì M(x)>=3 với mọi x)
do đó đa thức M(x) không có nghiệm

a)

`P(x)=7x^3+(4x^2-3x^2)-x+5=7x^3+x^2-x+5`

`Q(x)=-7x^3-x^2+2x+(6-8)=-7x^3-x^2+2x-2`

b)

`P(x)+Q(x) = 7x^3+x^2-x+5-7x^3-x^2+2x-2`

`=(7x^3-7x^3)+(x^2-x^2)+(2x-x)+(5-2)`

`=x+3`

`P(x)-Q(x)=7x^3+x^2-x+5-(-7x^3-x^2+2x-2)`

`= 7x^3+x^2-x+5+7x^3+x^2-2x+2`

`=(7x^3+7x^3)+(x^2+x^2)-(x+2x)+(5+2)`

`=14x^3+2x^2-3x+7`

c) `A(x) = P(x)+Q(x)=x+3`

`A(x)=0 <=> x+3=0 <=>x=-3`.