Question

50 БАЛЛОВ Помогите с любым номером, в первом задании при а=2, n=-1, c=-3

Answer 1

1.
$\frac{n-c}{a+n} - \frac{an-n^2}{a^2-ac} * \frac{a^2-c^2}{a^2-n^2} +11,5n= \\ \\ = \frac{n-c}{a+n} - \frac{n(a-n)}{a(a-c)} * \frac{(a-c)(a+c)}{(a-n)(a+n)} +11.5n = \\ \\ = \frac{n-c}{a+n} - \frac{n(a+c)}{a(a+n)}+ 11.5n= \frac{a(n-c) -n(a+c)}{a(a+n)} +11.5n= \\ \\ = \frac{an-ac-an -cn}{a(a+n)} +11.5n= \frac{-ac-cn}{a(a+n)} +11.5n= \frac{-c(a+n)}{a(a+n)} +11.5n= \\ \\ = \frac{-c}{a} +11.5n= - \frac{c}{a} +11.5n$

при а = 2 ; n=-1 ; с = -3
$- \frac{-3}{2} + 11.5n= -(-1,5) + 11.5*(-1)= 1,5 -11.5 = -10$

2.
$\frac{2x-y}{xy} - \frac{1}{x+y} *( \frac{x}{y} - \frac{y}{x} )= \frac{1}{y} \\ \\ \frac{2x-y}{xy} - \frac{1}{x+y} *\frac{x^2-y^2}{xy}= \frac{1}{y} \\ \\ \frac{2x-y}{xy} - \frac{1*(x-y)(x+y)}{(x+y)*xy} = \frac{1}{y} \\ \\ \frac{2x-y}{xy} - \frac{x-y}{xy}= \frac{1}{y} \\ \\ \frac{2x-y-(x-y)}{xy} = \frac{1}{y} \\ \\ \frac{2x-y-x+y}{xy} = \frac{1}{y} \\ \\ \frac{x}{xy} = \frac{1}{y} \\ \\ \frac{1}{y} = \frac{1}{y}$
тождество доказано.

3. не видно знака ( х + 1 ???  1/(х+1)) ...