Question

Помогите решить систему уравнений

Answer 1

x=3 y=1


$left { {{log_ {sqrt{2}} (x-y)=2 } atop {2^x*5^{x-2y}=40 }} right. \ \ log_ {sqrt{2}} (x-y)=2 \ x-y= (sqrt{2} )^2=2 \ \ left { {{x-y=2 } atop {2^x*5^{x-2y}=40 }} right. \ left { {{x=2 +y} atop {2^{2+y}*5^{(2+y)-2y}=40 }} right. \ \ 2^{2+y}*5^{(2+y)-2y}=40 \ 2^{2+y}*5^{2-y}=40\ 2^2*2^y*5^2* frac{1}{5^y} =40 \ 4*2^y*25* frac{1}{5^y} =40 \ 100 *frac{2^y}{5^y} =40 \ (frac{2}{5})^y = frac{40}{100} \ 0.4^y=0.4 \ y=1 \ \ left { {{x=2+y} atop {y=1}} right.$
$\ left { {{x=2+1} atop {y=1}} right. \ left { {{x=3} atop {y=1}} right.$