Question

Решите пожалуйста номер 375. Срочно нужна ваша помощь

Answer 1

1)
$x- \frac{ x^{2} +4 y^{2}}{x+2y} +2y=\frac{ x(x+2y)-(x^{2} +4 y^{2})+2y(x+2y)}{x+2y} = \\ \frac{ x^{2} +2xy-x^{2} -4 y^{2}+2xy+4y^{2}}{x+2y} = \frac{4xy}{x+2y}$

2)Заметим вначале, что в область допустимых значений х не входят числа -1, 0 и 1.
$\frac{3}{2 x^{2} +2x} + \frac{2x-1}{ x^{2} -1} - \frac{2}{x} = \frac{3}{2 x(x+1)} + \frac{2x-1}{(x+1)(x-1)} - \frac{2}{x} = \\ \frac{3(x-1)}{2x(x+1)(x-1)} + \frac{2x(2x-1)}{2x(x+1)(x-1)} - \frac{4( x^{2} -1)}{2x(x+1)(x-1)} = \\ \frac{3x-3+2 x^{2} -2x-4 x^{2} +4}{2x(x+1)(x-1)} = \frac{-2 x^{2}+x+1}{2x(x+1)(x-1)} = \frac{-2 x^{2}+2x-x+1}{2x(x+1)(x-1)} = \\ \frac{-2 x(x-1)-(x-1)}{2x(x+1)(x-1)} =\frac{-(2x+1)(x-1)}{2x(x+1)(x-1)} =\frac{-(2x+1)}{2x(x+1)} =-\frac{2x+1}{2x(x+1)}$

3)
$\frac{2b}{2b+3} + \frac{5}{3-2b} - \frac{4b^{2}+9 }{4b^{2}-9} = \frac{2b}{2b+3} + \frac{5}{3-2b} + \frac{4b^{2}+9 }{(2b+3)(3-2b)} = \\ \frac{2b(3-2b)+5(2b+3)+4b^{2}+9 }{(2b+3)(3-2b)} = \frac{6b-4b^{2} +10b+15+4b^{2}+9 }{(2b+3)(3-2b)} = \frac{16b+24}{9-4b^{2} }$

4)
$\frac{5y}{ x^{2} -25 y^{2} } + \frac{y}{5 y^{2} -xy} = \frac{5y}{ (x+5y)(x-5y) } - \frac{y}{y(x-5y)} = \frac{5y^{2}-y(x+5y)}{ y(x+5y)(x-5y) } = \\ \frac{5y^{2}-xy-5y^{2}}{ y(x+5y)(x-5y) } =\frac{-xy}{ y(x+5y)(x-5y) } =\frac{xy}{ y(x+5y)(5y-x) }$