a) G(x) = 2x⁵-4x⁴-10x³+3x²-4x-8

      H(x) = x⁵-2x⁴-5x³+x²+7x-4

b) G(x)+H(x)=3x⁵-6x⁴-15x³+4x²+3x-12

    G(x)-H(x) =x⁵-2x⁴-5x³+2x²-11x-4

c) G(x) = 2H(x)

2x⁵-4x⁴-10x³+3x²-4x-8=2( x⁵-2x⁴-5x³+x²+7x-4)

2x⁵-4x⁴-10x³+3x²-4x-8-2( x⁵-2x⁴-5x³+x²+7x-4)=0

2x⁵-4x⁴-10x³+3x²-4x-8-2x⁵+4x⁴⁺10x³-2x²-14x+8=0

x²-18x=0

x(x-18)=0

x=0 hoặc x-18=0

                x=18

a: \(A\left(x\right)=2x^4-x^3+3x^2+9x-2\)

\(B\left(x\right)=2x^4-5x^3-x+9\)

\(C\left(x\right)=x^4+4x^2+5\)

A(x): bậc 4; hệ số cao nhất là 2; hệ số tự do là -2

B(x): bậc 4; hệ số cao nhất là 4; hệ số tự do là 9

b: M(x)=A(x)+B(x)=4x^4-6x^3+3x^2+8x+7

N(x)=B(x)-A(x)=-4x^3-3x^2-10x+11

c: Q(x)=-N(x)=4x^3+3x^2+10x-11

`a)`

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

          `=(2x^4-5x^4)+x^3-4x-4`

          `=-3x^4+x^3-4x-4`

`@P(x)-Q(x)=-2x^4+x^3+2x^2-4x-1+3x^4-x^3+4x+4`

                  `=x^4+2x^2+3`

______________________________________

`b)P(x)-Q(x)=0`

`=>x^4+2x^2+3=0`

`=>(x^2)^2+2x^2+1+2=0`

`=>(x^2+1)^2+2=0`

`=>(x^2+1)^2=-2` (Vô lí vì `(x^2+1)^2 >= 0` mà `-2 < 0`)

Vậy đa thức `P(x)-Q(x)` không có nghiệm

a)\(P\left(x\right)=M\left(x\right)+N\left(x\right)\)

\(P\left(x\right)=x^4+3x-\dfrac{1}{9}-x+3x^4+2x^2+8x-2x^3+2x^3+\dfrac{2}{3}+4x-4x^4-\dfrac{1}{3}\)

\(P\left(x\right)=2x^2+\dfrac{2}{9}+14x\)

bài 3:

a) f(x)= x²+2x⁴-2x³+x²+5x⁴+4x³-x+5

= (2x⁴+5x⁴)+(4x³-2x³)+(x²+x²)-x+5

= 7x⁴+2x³+2x²-x+5

g(x)= -2x²+8x⁴+x-x⁴-3x³+3x²+5+4x³

=(8x⁴-x⁴)+(4x³-3x³)+(3x²-2x²)+x+5

= 7x⁴+x³+x²+x+5

b) h(x)=f(x)-g(x)

=(7x⁴+2x³+2x²-x+5)-(7x⁴+x³+x²+x+5)

=7x⁴+2x³+2x²-x+5-7x⁴-x³-x²-x-5

=(7x⁴-7x⁴)+(2x³-x³)+(2x²-x²)-(x+x)+(5-5)

=x³+x²-2x

Bài 4:

a) f(x)=5x⁴+x³-x+11+x⁴-5x³

=(5x⁴+x⁴)+(x³-5x³)-x+11

=6x⁴-4x³-x+11

g(x)=2x³+3x⁴+9-4x³+2x⁴-x

=(3x⁴+2x⁴)+(2x³-4x³)-x+9

=5x⁴-2x³-x+9

b) h(x)=f(x)-g(x)

=(6x⁴-4x³-x+11)-(5x⁴-2x³-x+9)

=6x⁴-4x³-x+11-5x⁴-2x³-x+9

=(6x⁴-5x⁴)-(4x³+2x³)-(x+x)+(11+9)

= x⁴-6x³-2x+20

c) Với x = -2

Ta có: h(-2)=(-2)⁴-6.(-2)³-2.(-2)+20=88\(\ne\)0

Vậy x = -2 không phải là nghiệm của đa thức h(x)

đúng thì tặng 1 tick cho mk nk các pn!!!

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

\(=9x^4+2x^2-x+5\)

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

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

Ta có: P(x)+Q(x)

\(=9x^4+2x^2-x+5-x^4-x^3-2x^2-2x-1\)

\(=8x^4-x^3-3x+4\)

Ta có: P(x)-Q(x)

\(=9x^4+2x^2-x+5+x^4+x^3+2x^2+2x+1\)

\(=10x^4+x^3+4x^2+x+6\)

a)\(M\left(x\right)=3x^4-x^3-2x^2+5x+7\)

\(N\left(x\right)=-3x^4+x^3+10x^2+x-7\)

b)\(A\left(x\right)=M\left(x\right)+N\left(x\right)\)

\(=>A\left(x\right)=3x^4-x^3-2x^2+5x+7-3x^4+x^3+10x^2+x-7\)

\(A\left(x\right)=8x^2+6x\)

\(B\left(x\right)=3x^4-x^3-2x^2+5x+7+3x^4-x^3-10x^2-x+7\)

\(B\left(x\right)=6x^4-2x^3-12x^2+x+14\)

c)cho A(x) = 0

\(=>8x^2+6x=0=>x\left(8x+6\right)=0=>\left[{}\begin{matrix}x=0\\8x=-6\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{4}\end{matrix}\right.\)