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

Bậc là 5

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

Bậc là 5

b: H(x)=P(x)+Q(x)

\(=5x^5-4x^4-2x^3+4x^2+3x+6-5x^5+4x^4+2x^3-4x^2+7x+\dfrac{1}{4}\)

=10x+6,25

c: Để H(x)=0 thì 10x+6,25=0

hay x=-0,625

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

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

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

c: Bậc của H(x) là 3

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

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

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

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

c) Bậc : 3

d)\(H\left(3\right)=-4.3^3+2.3^2+4.3=-4.27+2.9+12=-108+18+12=-78\)

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

\(H\left(-3\right)=-4.\left(-27\right)+2.9-12=108+18-12=114\)

\(H\left(\dfrac{1}{3}\right)=-4.\left(\dfrac{1}{3}\right)^3+2.\left(\dfrac{1}{3}\right)^2+\dfrac{4.1}{3}=-\dfrac{4.1}{27}+\dfrac{2.1}{9}+\dfrac{4}{3}\)

\(H\left(\dfrac{1}{3}\right)=-\dfrac{4}{27}+\dfrac{6}{27}+\dfrac{36}{27}=\dfrac{38}{27}\)

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,

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

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

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

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

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

a) Sắp xếp theo lũy thừa giảm dần

P(x)=x^5−3x^2+7x^4−9x^3+x^2−1/4x

=x^5+7x^4−9x^3−3x^2+x^2−1/4x

=x^5+7x^4−9x^3−2x^2−1/4x

Q(x)=5x^4−x^5+x^2−2x^3+3x^2−1/4

=−x^5+5x^4−2x^3+x^2+3x^2−1/4

=−x^5+5x^4−2x^3+4x^2−1/4

b)

P(x)+Q(x)

=(x^5+7x^4−9x^3−2x^2−1/4^x)+(−x^5+5x^4−2x^3+4x^2−1/4)

=x^5+7x^4−9x^3−2x^2−1/4x−x^5+5x^4−2x^3+4x^2−1/4

=(x^5−x^5)+(7x^4+5x^4)+(−9x^3−2x^3)+(−2x^2+4x^2)−1/4x−1/4

=12x^4−11x^3+2x^2−1/4x−1/4

P(x)−Q(x)

=(x^5+7x^4−9x^3−2x^2−1/4x)−(−x^5+5x^4−2x^3+4x^2−1/4)

=x^5+7x^4−9x^3−2x^2−1/4x+x^5−5x^4+2x^3−4x^2+1/4

=(x^5+x^5)+(7x^4−5x^4)+(−9x^3+2x^3)+(−2x^2−4x^2)−1/4x+1/4

=2x5+2x4−7x3−6x2−1/4x−1/4

c) Ta có

P(0)=0^5+7.0^4−9.0^3−2.0^2−1/4.0

⇒x=0là nghiệm của P(x).

Q(0)=−0^5+5.0^4−2.0^3+4.0^2−1/4=−1/4≠0

⇒x=0không phải là nghiệm của Q(x).

a) P(x) = 2x³ - 2x - x² - x³ + 3x + 2

=> P(x) = (2x³ - x³) + (-2x + 3x) - x² + 2

=> P(x) = x³ + x - x² + 2

Sắp xếp : P(x) = x³ - x² + x + 2

Q(x) = -4x³ + 5x² - 3x + 4x + 3x³ - 4x² + 1

=> Q(x) = (-4x³ + 3x³) + (5x² - 4x²) + (-3x + 4x) + 1

=> Q(x) = -x³ + x² + x + 1

Sắp xếp : Q(x) = -x³ + x² + x + 1

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

=> H(x) = (x³ + x - x² + 2) + (-x³ + x² + x + 1)

=> H(x) = x³ + x - x² + 2 - x³ + x² +x + 1

=> H(x) = (x³ - x³) + (x + x) + (-x² + x²) + (2 + 1)

=> H(x) = 2x + 3

K(x) = P(x) - Q(x)

=> K(x) = (x³ + x - x² + 2) - (-x³ + x² + x + 1)

=> K(x) = x³ + x - x² + 2 + x³ - x² - x - 1

=> K(x) = (x³ + x³) + (x - x) + (-x² - x²) + (2 - 1)

=> K(x) = 2x³ - 2x² + 1

c) Q(2) = -2³ + 2² + 2 + 1 = -8 + 4 + 2 + 1 = -1( m k bt (-2)³ hay -2³ nx nên thông cảm))

P(-1) = (-1)³ - (-1)² + (-1) + 2 = -1 - 1 - 1 + 2 = -1

d) Để H(x) có nghiệm => 2x + 3 = 0 => 2x = -3 => \(x=-\frac{3}{2}\)

Vậy x = -3/2 là nghiệm của đa thức H(x)

P/s : K chắc :))

a) Mình làm tắt

P(x) = x³ - x² + x + 2

Q(x) = -x³ + x² + x + 1

b) H(x) = P(x) + Q(x) 

            =  x³ - x² + x + 2 - x³ + x² + x + 1

            = 2x + 3

K(x) = P(x) - Q(x)

        = x³ - x² + x + 2 - ( -x³ + x² + x + 1 )

        = x³ - x² + x + 2 + x³ - x² - x - 1

        = 2x³ - 2x² + 1

c) Q(2) = -(2)³ + 2² + 2 + 1 = -8 + 4 + 2 + 1 = -1

P(-1) =  1³ - 1² + 1 + 2 = 1 - 1 + 1 + 2 = 3

d) H(x) = 2x + 3

H(x) = 0 <=> 2x + 3 = 0

              <=> 2x = -3

              <=> = -3/2

Vậy nghiệm của H(x) = -3/2

a, Ta có : 

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

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

P/S :  Sắp xếp luôn r đấy.

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

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

c, Tự lm 

d, Đặt  \(2x+3=0\Leftrightarrow x=-\frac{3}{2}\)

Vậy nghiệm của đa thức là -3/2