a,

 \(Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\\ =\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2+\left(-3x+3x\right)+\left(\dfrac{2}{3}+1\right)\\ =3x^4+0+2x^2+0+\dfrac{5}{3}\\ =3x^4+2x^2+\dfrac{5}{3}\)

b, Ta có

\(\left\{{}\begin{matrix}x^4\ge0\\x^2\ge0\end{matrix}\right.\\ \Rightarrow3x^4+2x^2\ge0\\ \Rightarrow3x^4+2x^2+\dfrac{5}{3}\ge\dfrac{5}{3}>0\)

\(\Rightarrow Q\left(x\right)\) lớn hẳn hơn 0

\(\Rightarrow Q\left(x\right)\) vô nghiệm 

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

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

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

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

b: P(x)+Q(x)

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

\(=x^4+2\)

P(x)-Q(x)

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

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

a: P(x)=5x^3+3x^2-2x-5

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

b: P(x)-Q(x)=x^2-9

P(x)+Q(x)=10x^3+5x^2-4x-1

c: P(x)-Q(x)=0

=>x^2-9=0

=>x=3; x=-3

d: C=A*B=-7/2x^6y^4

a: P(x)=6x^4+5x^3-3x^2+5x-10

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

b: P(x)+Q(x)

=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4

=11x^4+10x^3-x^2+x-6

P(x)-Q(x)

=6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4

=x^4-5x^2+9x-14

a: P(x)=6x^4+5x^3-3x^2+5x-10

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

b: P(x)+Q(x)

=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4

=11x^4+10x^3-x^2+x-6

P(x)-Q(x) =6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4

=x^4-5x^2+9x-14

cái Q(x)=\(5x^2-4x^3-2x+7\)

mik ghi nhầm xin lổy đc chx 

a) \(P\left(x\right)=6x^3-3x^2+5x-1\)

\(Q\left(x\right)=5x^2-4x^2-2x+7=\left(5x^2-4x^2\right)-2x+7=x^2-2x+7\) ( Kết quả này cũng giống như sắp xếp nhé)

b) \(P\left(x\right)+Q\left(x\right)=6x^3+5x-3x^2-1+\left(x^2-2x+7\right)\)

\(=6x^3+5x-3x^2-1+x^2-2x+7\)

\(=6x^3+\left(5x-2x\right)+\left(-3x^2+x^2\right)+\left(-1+7\right)\)

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

a)\(P\left(x\right)=2.2+2x-6.2+4.3+2-3x\)

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

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

\(=\left(-x\right)+18\)

\(Q\left(x\right)=3-2.4+3x+2.4+3.3-x\)

\(=3-8+3x+8+9-x\)

\(=\left(3x-x\right)+\left(3-8+8+9\right)\)

\(=2x+12\)

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

\(\Rightarrow C\left(x\right)=\left(-x+18\right)+\left(2x+12\right)\)

\(C\left(x\right)=-x+18+2x+12\)

\(C\left(x\right)=\left(-x+2x\right)+\left(18+12\right)\)

\(C\left(x\right)=x+26\)

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

\(\Rightarrow D\left(x\right)=\left(2x+12\right)-\left(-x+18\right)=\)

\(D\left(x\right)=2x+12+x+18\)

\(D\left(x\right)=\left(2x+x\right)+\left(12+18\right)\)

\(D\left(x\right)=3x+26\)

Có \(3x\) luôn \(\ge0\) và \(\le0\) với mọi x.

Lại có 26>0; \(26⋮̸3\)

\(\Rightarrow\)D(x) vô nghiệm

`@` `\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) = 7x³ - x² + 5x - 2x³ +6 - 8x

=5x^3-x^2-3x+6

 Q(x) = -2x + x³ - 4x² + 3 - 5x²

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

b)

P(x) - Q(x)=4^3+8x^2-x-3

P(x) + Q(x)=6^3-10x^2-5x+9

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

Vậy Q(x) không có nghiệm