Во вложения. Кто сможет - тот решит.

Question 1

Во вложения.

Кто сможет - тот решит.

Question 2

Решите задачу:

$\frac{48p^3q^4}{36p^2q^3} = \frac{4pq}{3} \\\ \frac{x^2+6x+9}{x^2-9} = \frac{(x+3)^2}{(x-3)(x+3)} =\frac{x+3}{x-3}$
$\frac{10x^2}{9y^2} = \frac{40x^3}{36xy^2} \\\\ \frac{8}{12xy} = \frac{24y}{36xy^2}$
$\frac{b+2}{4b} = \frac{(b+2)(b-2)}{4b(b-2)} \\\\ \frac{4b+5}{4b-8} = \frac{b(4b+5)}{4b(b-2)}$
$\frac{2c}{c+d} = \frac{2c(d-c)}{(c+d)(d-c)} \\\\ \frac{3d}{d-c} = \frac{3d(c+d)}{(c+d)(d-c)}$
$\frac{5t}{t^2-25} = \frac{5t}{(t-5)(t+5)} = \frac{5t(t-5)}{(t-5)^2(t+5)} \\\ \frac{t+5}{(t-5)^2} =\frac{(t+5)^2}{(t-5)^2(t+5)}$

Question 3

1. а) $\frac{48p^3q^4}{36p^2q^3} = \frac{4pq}{3}$
б) $\frac{x^2+6x+9}{x^2-9} = \frac{(x+3)^2}{(x-3)(x+3)} = \frac{x+3}{x-3}$

2. а) $\frac{10x^2}{9y^2}$ и $\frac{8}{12xy}$
$\frac{10x^2}{9y^2} = \frac{10x^2*x}{9y^2*x} = \frac{10x^3}{9y^2x} \\ \frac{8}{12xy} = \frac{2}{3xy} = \frac{2*3y}{3xy*3y} = \frac{6y}{9y^2x}$

б) $\frac{b+2}{4b}$ и $\frac{4b+5}{4b-8}$
$\frac{b+2}{4b} = \frac{(b+2)(b-2)}{4b(b-2)} = \frac{b^2-4}{4b^2-8b} \\ \frac{4b+5}{4b-8} = \frac{(4b+5)b}{(4b-8)b} = \frac{4b^2+5b}{4b^2-8b}$

в) $\frac{2c}{c+d}$ и $\frac{3d}{d-c}$
$\frac{2c}{c+d} = \frac{2c(d-c)}{(c+d)(c-d)} = \frac{2cd-2c^2}{d^2-c^2} \\ \frac{3d}{d-c} = \frac{3d(c+d)}{(d-c)(c+d)} = \frac{3cd+3d^2}{d^2-c^2}$

г) $\frac{5t}{t^2-25}$ и $\frac{t+5}{(t-5)^2}$
$\frac{5t}{t^25} = \frac{5t(t-5)}{(t^2-25)(t-5)}= \frac{5t^2-25t}{t^3-5t^2-25t+125} \\ \frac{t+5}{(t-5)^2} = \frac{(t+5)^2}{(t-5)^2(t+5)} = \frac{t^2+10t+25}{t^3-5t^2-25t+125}$