c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)

\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)

\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)

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

c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

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

\(=\left(x-1\right)^2\)

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)

b: \(\left(x^2+10x+8\right)^2-\left(8x+4\right)\left(x^2+8x+7\right)\)

\(=\left\lbrack\left(x^2+8x+7\right)+\left(2x+1\right)\right\rbrack^2-4\left(2x+1\right)\left(x^2+8x+7\right)\)

\(=\left(x^2+8x+7\right)^2+2\left(x^2+8x+7\right)\left(2x+1\right)+\left(2x+1\right)^2-4\left(2x+1\right)\left(x^2+8x+7\right)\)

\(=\left(x^2+8x+7\right)^2-2\left(x^2+8x+7\right)\left(2x+1\right)+\left(2x+1\right)^2\)

\(=\left(x^2+8x+7-2x-1\right)^2=\left(x^2+6x+6\right)^2\)

d: \(B=x^4+4x^3+8x^2+8x+4\)

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

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

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

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

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

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

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

Thu gọn và sắp xếp phải k ạ?

`F(x)= (x^4-x^4)+(5x^2-8x^2)-4x+x^5+3+2x^3+2`

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

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

\(F\left(x\right)=x^4+5x^2-4x+x^5-x^4-8x^2+3+2x^3+2\)

\(F\left(x\right)=x^5+\left(x^4-x^4\right)+2x^3+\left(5x^2-8x^2\right)-4x+\left(3+2\right)\)

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

f (x) = 3x² + 2x³ - 6x - 2

bậc của đa thức là: 3

g(x) = 5x² + 9 - 2x³ - 3x² - 4x + 2x³ - 2

g(x) = ( 5x² - 3x² ) + ( 9 -2) + ( - 2x³ + 2x³ ) - 4x

g(x) = 2x² + 7 - 4x

bậc của đa thức là : 2

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

Q(x)=-5x^3-10x^2+6x+11

M(x)=x^3-14x^2+10x+9

b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)

=10x^4-11x^3-5x^2-15x+21

= 5/3 - (8x-1)/3 = (4x + 2)/5 - 25/5

=> (5-8x+1)/3 = (4x + 2 - 25)/5

=> (6-8x)*5 = (4x-23)*3

=> 30 - 40x = 12 x - 69

=> 52x = 99

x = 90/52