Respuesta :
[tex]\frac{x^{2} - 5x + 6}{15xy^{2}} \div \frac{2x^{2} - 7x + 3}{5x^{2}y}[/tex]
[tex]\frac{x^{2} - 5x + 6}{15xy^{2}} \times \frac{5x^{2}y}{2x^{2} - 7x + 3}[/tex]
[tex]\frac{(x - 3)(x - 2)}{3y} \times \frac{x}{(2x - 1)(x - 3)}[/tex]
[tex]\frac{x - 2}{3y} \times \frac{x}{2x - 1}[/tex]
[tex]\frac{x^{2} - 2x}{6xy - 3y}[/tex]
The answer is C.
[tex]\frac{x^{2} - 5x + 6}{15xy^{2}} \times \frac{5x^{2}y}{2x^{2} - 7x + 3}[/tex]
[tex]\frac{(x - 3)(x - 2)}{3y} \times \frac{x}{(2x - 1)(x - 3)}[/tex]
[tex]\frac{x - 2}{3y} \times \frac{x}{2x - 1}[/tex]
[tex]\frac{x^{2} - 2x}{6xy - 3y}[/tex]
The answer is C.
Answer:
C is the correct option.
Step-by-step explanation:
We have been given the expression
[tex][/tex]
When we flip the denominator, the division sign gets changed to multiplication. The rule is shown below
[tex]\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\times\frac{d}{c}[/tex]
Using this rule, we have
[tex]\frac{x^2-5x+6}{15xy^2}\times\frac{5x^2y}{2x^2-7x+3}[/tex]
Now, we can factor the quadratic expressions as
[tex]x^2-5x+6=x^2-3x-2x+6=(x-3)(x-2)[/tex]
[tex]2x^2-7x+3=2x^2-6x-x=(x-3)(2x-1)[/tex]
On plugging these values, we get
[tex]\frac{(x-3)(x-2)}{3\cdot5\cdot xy^2}\times\frac{5x^2y}{(x-3)(2x-1)}[/tex]
On cancelling the common terms in numerator and denominator, we get
[tex]\frac{x(x-2)}{3y(2x-1)}[/tex]
Hence, C is the correct option.