Question

Решите 3 задания пожалуйста

Answer 1

Решение:

2.

$\Big (\dfrac{x}{x-y} +\dfrac{2xy}{x^2-2xy+y^2}\Big ):\Big (\dfrac{x-y}{2x} -1\Big ) =\\ \\= \Big (\dfrac{x}{x-y} +\dfrac{2xy}{(x-y)^2}\Big ):\Big (\dfrac{x-y-2x}{2x} \Big ) =\\ \\= \Big (\dfrac{x(x-y)}{(x-y)^2} +\dfrac{2xy}{(x-y)^2}\Big ):\dfrac{-(x+y)}{2x} =\\ \\=\Big (\dfrac{x^2-xy+2xy}{(x-y)^2} \Big ):\dfrac{-(x+y)}{2x} =\\ \\=\dfrac{x^2+xy}{(x-y)^2}:\dfrac{-(x+y)}{2x} =\dfrac{x(x+y)}{(x-y)^2}:\dfrac{-(x+y)}{2x} =$

$=\dfrac{x(x+y)}{(x-y)^2}\cdot \dfrac{2x}{-(x+y)} =-\dfrac{2x^2}{(x-y)^2} .$

При х = -2 и у = -1

$-\dfrac{2\cdot(-2)^2}{(-2+1)^2} =-8.$

3.

Пусть функция имеет вид:

$y = \dfrac{k}{x}$

При х = 7 у = 49

тогда

k = 7 · 49 = 343

и

$y = \dfrac{343}{x}$.

4.

$(\sqrt{6} + \sqrt{3} )\cdot \sqrt{12} - 2\sqrt{6} \cdot \sqrt{3} = \sqrt{6}\cdot \sqrt{12} + \sqrt{3}\cdot \sqrt{12} - 2 \sqrt{6\cdot 3} = \\ \\=\sqrt{36\cdot 2} + \sqrt{36} - 2\sqrt{9\cdot2 } =6\sqrt{2} +6 -6\sqrt{2} =6.$