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

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

Ta có: \(g\left(x\right)=2x^3+10x-1-7x^2-15x+10x^2\)

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

b) Ta có: f(x)+g(x)

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

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

Ta có: f(x)-g(x)

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

\(=x^4-2x^3-6x^2+7x-4\)

Thank bn

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!!!

giải câu c ở bài 3 với

`Answer:`

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

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

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

\(g\left(x\right)=-2x^3+10x-1-7x^2+x^4-15x+10x^2\)

\(=x^4-2x^3+\left(-7x^2+10x^2\right)+\left(10x-15x\right)-1\)

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

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

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

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

a)  \(F\left(x\right)=-x^5-7x^4-2x^3+x^2+4x+9\)

\(G\left(x\right)=x^5+7x^4+2x^3+2x^2-3x-9\)

b)   \(F\left(x\right)+G\left(x\right)=x+3x^2\)

F(x) + G(x) = \(9-x^5+4x-2x^3+x^2-7x^4-x^5+9-2x^2-7x^4-2x^3+3x\)

=\(18-2x^5+7x-4x^3-x^2-14x^4\)

c) F(x) + G(x) = \(x+3x^2=0\Rightarrow x.\left(1+3x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\1+3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{3}\end{cases}}\)

Ta có:

* \(f\left(x\right)=15-4x^3+2x-x^3+x^2-10\)

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

*\(g\left(x\right)=4x^3+6x^2-5x+5-9x^3+7x\)

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

a) \(f\left(x\right)-g\left(x\right)=\)\(-5x^3+x^2+2x+5-\left(-5x^3+6x^2+2x+5\right)\)

\(=x^2-6x^2\)

\(=-5x^2\)

b) Ta có: \(f\left(x\right)-g\left(x\right)=-5x^2\)(từ câu a)

\(\Rightarrow-5x^2=-125\)

\(\Rightarrow x^2=25\)\(\Rightarrow\orbr{\begin{cases}x=-5\\x=5\end{cases}}\)

a; F(x) = 4x^2 - 7x^2 + 4x - 5x^4 - x^2 + 6x^3 + 5x^4 - 5

F(x) = (4x^2 - 7x^2 - x^2) +4x +(-5x^4 + 5x^4) + 6x^3 - 5

F(x) = -4x^2 + 4x + 0 + 6x^3 - 5

F(x) = 6x^3 -4x^2 + 4x - 5

b; Bậc của đa thức là: 3

Hệ số tự do là - 5

Hệ số cao nhất là: 6

c; F(-1) = 6.(-1)^3 - 4(-1)^2 + 4.(-1) - 5

F(-1) = 6.(-1) - 4 - 4 - 5

F(-1) = - 6 - 4 - 4 - 5

F(-1) = -10 - 4 - 5

F(-1) = -14 - 5

F(-1) = - 19

F(0) = 6.0^3 - 4.0^2 + 4.0 - 5

F(0) = 0 - 0 - 0 - 5

F(0) = - 5

F(0,5) = 6.(0,5)^3 - 4.(0,5)^2 + 4.(0,5) - 5

F(0,5) = 6.0,125 - 4.0,25 + 2 - 5

F(0,5) = 0,75 - 1 + 2 - 5

F(0,5) = -0,25 + 2 - 5

F(0,5) = 1,75 - 5

F(0,5) = - 3,25

F(1) = 6.(1)^3 - 4.(1)^2 + 4.(1) - 5

F(1) = 6 - 4 + 4 - 5

F(1) = 2+ 4 - 5

F(1) = 6 - 5

F(1) = 1

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

Bài 1:

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

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

\(=2x-5\)

Bài 1: 

b) 

\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)

\(P\left(3\right)=2\cdot3-5=6-5=1\)