Question

по алгебре задали карточку с заданиями сидим уже 2 часа не можем решить

Answer 1

Номер 1.

$1) \frac{3x^{2} }{10} *\frac{5}{x} =\frac{3x*1}{2*1} =\frac{3x}{2}$

$2) (x+1)*\frac{5x-3}{4x+4} =\frac{x+1}{1} *\frac{5x-3}{4(x+1)} =\frac{5x-3}{4}$

$3) \frac{3x+9}{x-4} *\frac{4x-16}{x^{2} - 9 } = \frac{3(x+3)}{x-4} *\frac{4(x-4)}{(x-3)(x+3)} = \frac{3*4}{1*(x-3)} =\frac{12}{x-3}$

Номер 2.

$1) 2x:\frac{3x^{3} }{x-4} =\frac{2x}{1} *\frac{x-4}{3x^{3} } = \frac{2(x-4)}{3x^{2} } =\frac{2x-8}{3x^{2} }$

$2) \frac{5x+15}{3x-15} :\frac{3x+9}{x^{2} -25 } = \frac{5x+15}{3x-15} *\frac{x^{2}-25 }{3x+9} =\frac{5(x+3)}{3(x-5)} *\frac{(x-5)(x+5)}{3(x+3)} =\frac{5(x+5)}{3*3} =\frac{5x+25}{9}$

$3) \frac{7x-14}{(x-3)^{2} } : \frac{2x-4}{x^{2}-9 } =\frac{7x-14}{(x-3)^{2} } *\frac{x^{2} -9}{2x-4} =\frac{7(x-2)}{(x-3)^{2} } *\frac{(x-3)(x+3)}{2(x-2)} =\frac{7(x+3)}{(x-3)2} =\frac{7x+21}{2x-6}$