Question

Làm phép chia

a, \(3x^3y^2:x^2\)

b, \(\left(x^5+4x^3-6x^2\right):4x^2\)

c, \(\left(x^3-8\right):\left(x^2+2x+4\right)\)

d, \(\left(3x^2-6x\right):\left(2-x\right)\)

e, \(\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\)

Answer 1

\(\text{a) }3x^2y^2:x^2=3y^2\)

\(\text{b) }\left(x^5+4x^3-6x^2\right):4x^2\\ =\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

\(\text{c) }\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

\(\text{d) }\left(3x^2-6x\right):\left(2-x\right)\\ =3x\left(x-2\right):\left(2-x\right)\\ =-3x\left(2-x\right):\left(2-x\right)\\ =-3x\)

\(\text{e) }\left(x^3+2x^2-2x-1\right):\left(x^2+3x+1\right)\\ =\left(x^3+3x^2-x^2+x-3x-1\right):\left(x^2+3x+1\right)\\ =\left[\left(x^3+3x^2+x\right)-\left(x^2+3x+1\right)\right]:\left(x^2+3x+1\right)\\ =\left[x\left(x^2+3x+1\right)-\left(x^2+3x-1\right)\right]:\left(x^2+3x+1\right)\\ =\left(x-1\right)\left(x^2+3x+1\right):\left(x^2+3x+1\right)\\ =x-1\)

Answer 2

a) 3x²y² : x² = 3y²

b)( x⁵ + 4x³ - 6x² ) : 4x²

=\(\dfrac{1}{4}\)x³+ x - \(\dfrac{3}{2}\)