Question

Answer 1

Câu 3: 

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

\(=14x+6\)

b: ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)

\(=2x^2+14x-3x^2-3x\)

\(=-x^2+11x\)

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

\(=4x^2-12-x^3+3x+x^3-1\)

\(=4x^2+3x-13\)

d: Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2\)

\(=x^2-9-x^2+10x-25\)

=10x-34

Answer 2

Câu 2:

a. (-8x⁵ + 12x³ - 16x²) : 4x²

= (-8x⁵ : 4x²) + (12x³ : 4x²) - (16x² : 4x²)

= -2x³ + 3x - 4

b. (12x³y³ - 18x²y + 9xy²) : 6xy

= (12x³y³ : 6xy) - (18²y : 6xy) + (9xy² : 6xy)

= 6x²y² - 3x + \(\dfrac{3}{2}\)y

Answer 3

Câu 2: 

a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)

\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)

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

b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)

\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)

\(=2x^2y^2-3x+\dfrac{3}{2}y\)

c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)

\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)

\(=x^2-8x+3\)

d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)

\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)

\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)