Question

1) x-5y/4x-4y+ x+4y/5x-5y=2) 18x+12x³y/12x³+18x⁴​

Answer 1

Ответ:

Смотри

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Answer 2

Let's simplify each expression:

1. $\frac{x - 5y}{4x - 4y} + \frac{x + 4y}{5x - 5y}$

To add these fractions, you need a common denominator. In this case, the common denominator is $(4x - 4y)(5x - 5y)$.

$$\frac{(x - 5y)(5x - 5y)}{(4x - 4y)(5x - 5y)} + \frac{(x + 4y)(4x - 4y)}{(4x - 4y)(5x - 5y)}$$

Now, combine the numerators:

$$\frac{5x^2 - 10xy - 5xy + 25y^2 + 4x^2 + 16xy - 4y^2}{(4x - 4y)(5x - 5y)}$$

Combine like terms:

$$\frac{9x^2 + 2xy + 21y^2}{(4x - 4y)(5x - 5y)}$$

Factor out the common factor in the denominator:

$$\frac{9(x^2 + \frac{2}{9}xy + \frac{7}{3}y^2)}{4(x - y)5(x - y)}$$

Simplify further:

$$\frac{9(x + \frac{1}{3}y)(x + \frac{7}{3}y)}{20(x - y)^2}$$

So, $\frac{x - 5y}{4x - 4y} + \frac{x + 4y}{5x - 5y} = \frac{9(x + \frac{1}{3}y)(x + \frac{7}{3}y)}{20(x - y)^2}$

2. $\frac{18x + 12x^3y}{12x^3 + 18x^4}$

Factor out the common factor in the numerator:

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

Cancel the common factor in the numerator and denominator:

$$\frac{3 + 2x^2y}{x^2(2 + 3x)}$$

So, $\frac{18x + 12x^3y}{12x^3 + 18x^4} = \frac{3 + 2x^2y}{x^2(2 + 3x)}$