Question

Пару заданий нужно... (

Answer 1

1.
$8 ^{ \frac{1}{3} + log_{2}3} - 5 ^{log_{25} 9} = (2^{3})^{ \frac{1}{3} } * (2^{3})^{ log _{2}3 } - 5 ^{ \frac{1}{2}*log _{5}9 } =2 * 2 ^{log _{2}9 } -9 ^ \frac{1}{2}$
$2 * 27 - 3 = 54 -3 =51$

2.
$\frac{2}{x} + \frac{1}{y} = \frac{5}{3} ; x= \frac{-6y}{-5y+3} \\ 2*\frac{-6y}{-5y+3} - y = 5 \\ -2*6y+y*5y-3*3=-5*5y+15 \\ y ^{2} + 2y - 3 = 0 \\ D = 4-4*(-3)=4+12=16 \\ y_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{-2+4}{2}=1 \\ y _{2}= \frac{-b- \sqrt{D} }{2a} = \frac{-2-4}{2}=-3 \\ y=1 x= \frac{-6}{-5+3}=3 \\ y=-3 x= \frac{18}{18}=1 \\ solution: (3;1) (1;-3)$