# X^2-5xÃ·2x-3yÃ·x^2-25Ã·4x^2-9y^2?

Answers (1)

$\frac{x^2-5x}{2x-3y}\div \frac{x^2-25}{4x^2-9y^2}\\* \Rightarrow \because \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}\\* So,\\* \Rightarrow \frac{x^2-5x}{2x-3y}\times \frac{4x^2-9y^2}{x^2-25}\\* \Rightarrow \frac{x(x-5)\times (2x+3y)(2x-3y)}{(2x-3y)\times (x+5)(x-5)}\\*\Rightarrow \frac{x(2x+3y)}{(X+5)}=\frac{2x^2+3xy}{x+5}$

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