Question

Решите задание по алгебре 10-11 класс)

Answer 1

$left { {{left(frac{1}{3}right)^{2x-y}=27} atop {5^{3x-y}=frac{1}{25}}} right.$
$5^{3x-y}=frac{1}{25} \ 5^{3x-y}=25^{-1} \5^{3x-y}=left(5^2right)^{-1} \ 5^{3x-y}=5^{2left(-1right)} \ 3x-y=2left(-1right) \ 3x-y=-2 \ x=frac{-2+y}{3}$

$left(frac{1}{3}right)^{2cdot frac{-2+y}{3}-y}=27 \ \ left(frac{1}{3}right)^{2cdot frac{-2+y}{3}-y}=left(frac{1}{3}right)^{-3} \ \ 2cdot frac{-2+y}{3}-y=-3 \ \ frac{2left(-2+yright)}{3}-y=-3 \ \ 2left(-2+yright)-3y=-9 \ \ -4+2y-3y=-9 \ \ -4-y+4=-9+4 \ \ -y=-5 \ y=5 \ \ \ x=frac{-2+5}{3} \ x=1 \ \ OTB:y=5,:x=1$

$b) \ 12x+y=2 \ 64^{x-3y}=8 \ \ 12x+y=2 \ 12x=2-y \ frac{12x}{12}=frac{2}{12}-frac{y}{12} \ x=frac{2-y}{12} \ \ \ 64^{frac{2-y}{12}-3y}=8 \ left(8^2right)^{frac{2-y}{12}-3y}=8 \ 8^{2left(frac{2-y}{12}-3yright)}=8 \ 2left(frac{2-y}{12}-3yright)=1 \ frac{2left(frac{2-y}{12}-3yright)}{2}=frac{1}{2} \ frac{2-37y}{12}=frac{1}{2} \ frac{12left(2-37yright)}{12}=frac{1cdot :12}{2} \ 2-37y=6 \-37y=4 \ y=-frac{4}{37}$

$x=frac{2-left(-frac{4}{37}right)}{12}=frac{13}{74} \ \ y=-frac{4}{37},:x=frac{13}{74}$