Question

Hi.
Please i need help with these questions :

Simplify
1.
$\frac{ {d}^{2} - 9 }{ {d}^{2} - 7d + 12 }$
2.
$\frac{ {m}^{2} + 2mn + {n}^{2} }{ {m}^{2} - {n}^{2} }$
3.
$\frac{ {x}^{2} + xy - 6 {y}^{2} }{ {x}^{2} - 3xy + 2 {y}^{2} }$
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Answer 1

1)

$\frac{d^2 - 9}{d^2 - 7d + 12} = \frac{(d - 3)(d + 3)}{(d - 3)(d - 4)} = \frac{d + 3}{d - 4}$

2)

$\frac{m^2 + 2mn + n^2}{m^2 - n^2} = \frac{(m + n)(m + n)}{(m - n)(m + n)} = \frac{m + n}{m - n}$

3)

$\frac{x^2 + xy - 6y^2}{x^2 - 3xy + 2y^2} = \frac{(x + 3y)(x - 2y)}{(x - y)(x - 2y)} = \frac{x + 3y}{x - y}$

Answer 2

Solutions :

1. $\bf \dfrac{ {d}^{2} - 9 }{ {d}^{2} - 7d + 12 }$

$\tt : \implies \dfrac{ {d}^{2} - (3)^{2} }{ {d}^{2} - 3d - 4d + 12 }$

By using identinty a² - b² = (a + b)(a - b) in numerator and splitting method in denominator :

$\tt : \implies \dfrac{ (d+3)(d-3) }{ {d}^{2} - 3d - 4d + 12 }$

$\tt : \implies \dfrac{ (d+3)(d-3) }{d(d - 3) - 4(d - 3)}$

$\tt : \implies \dfrac{ (d+3)\cancel{(d-3)} }{\cancel{(d - 3)}(d - 4)}$

$\tt : \implies \dfrac{d+3}{d - 4}$

Hence, answer is $\boxed{\bf \dfrac{d+3}{d - 4} }$

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2. $\bf \dfrac{ {m}^{2} + 2mn + {n}^{2} }{ {m}^{2} - {n}^{2} }$

By using identinty a² + 2ab + b² = (a + b)² in numerator and a² - n² = (a + b)(a - b) in denominator :

$\tt : \implies \dfrac{(m+n)^{2}}{(m+n)(m-n)}$

$\tt : \implies \dfrac{\cancel{(m+n)}(m+n)}{\cancel{(m+n)}(m-n)}$

$\tt : \implies \dfrac{m+n}{m-n}$

Hence, answer is $\boxed{\bf \dfrac{m+n}{m-n}}$

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3. $\bf \dfrac{{x}^{2} + xy - 6{y}^{2}}{{x}^{2} - 3xy + 2{y}^{2}}$

By using splitting method in both numerator and denominator :

$\tt : \implies \dfrac{{x}^{2} + 3xy - 2xy - 6{y}^{2}}{{x}^{2} - xy - 2xy + 2{y}^{2}}$

$\tt : \implies \dfrac{x(x + 3y) - 2y(x + 3y)}{x(x - y) - 2y(x - y)}$

$\tt : \implies \dfrac{\cancel{(x-2y)}(x + 3y)}{\cancel{(x-2y)}(x - y)}$

$\tt : \implies \dfrac{x + 3y}{x - y}$

Hence, answer is $\boxed{\bf \dfrac{x + 3y}{x - y}}$