Question

упростить:
$( \frac{x - 3}{7x - 4} - \frac{x - 3}{x - 4} ) \div \frac{9x - 3x^{2} }{7x - 4} + \frac{ {x}^{2} - 14 }{4 - x}$
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Answer 1

Решение .

Упростить . Применяем формулу разности квадратов .

$\bf \displaystyle \Big(\frac{x-3}{7x-4}-\frac{x-3}{x-4}\Big):\frac{9x-3x^2}{7x-4}+\frac{x^2-14}{4-x}=\\\\\\=\frac{(x-3)(x-4)-(x-3)(7x-4)}{(x-4)(7x-4)}:\frac{3x\, (3-x)}{7x-4}+\frac{x^2-14}{4-x}=\\\\\\=\frac{(x-3)(x-4-7x+4)}{(x-4)(7x-4)}\cdot \frac{7x-4}{3x\, (3-x)}+\frac{x^2-14}{4-x}=\\\\\\=\frac{-6x\cdot (x-3)}{3x\, (x-4)(3-x)}-\frac{x^2-14}{x-4}==\frac{-2x\cdot (x-3)}{x\, (x-4)(3-x)}-\frac{x^2-14}{x-4}=\\\\\\=\frac{-2x\, (x-3)+x\, (x^2-14)(x-3)}{x\, (x-4)(3-x)}=\frac{(x-3)(-2x+x^3-14x)}{x\, (x-4)(3-x)}=$  

$\bf =\dfrac{(x-3)\cdot x\cdot (x^2-16)}{x\, (x-4)(3-x)}=\dfrac{(x-3)\cdot x\cdot (x-4)(x+4)}{-x(x-4)(x-3)}=-(x+4)=-x-4$