Question

Плизз хелп ми. Очень надо

Answer 1

$left { {x^{3}-y^{3}=7} atop {x-y=1}} right.\\:left { {{(x-y)(x^{2}+xy+y^{2})=7} atop {x-y=1}} right.\\left { {{x^{2}+xy+y^{2} =7} atop {x-y=1}} right.\\left { {{x=y+1} atop {(y+1)^{2}+y(y+1)+y^{2}=7 } right.\\left { {{y^{2}+2y+1+y^{2}+y+y^{2}=7} atop {x=y+1}} right.$

$left { {{3y^{2}+3y-6=0 } atop {x=y+1}} right.\\left { {{y^{2}+y-2=0} atop {x=y+1}} right.\\1)left { {{y=-2} atop {x=-2+1}} right.\\left { {{y=-2} atop {x=-1}} right.\\2)left { {{y=1} atop {x=1+1}} right.\\left { {{y=1} atop {x=2}} right.\\Otvet:(-1;-2),(2;1)$

$frac{2Cos^{2}alpha-1}{2SinalphaCosalpha}+frac{Sin3alpha -Sinalpha}{Cos3alpha+Cosalpha}=frac{Cos2alpha}{Sin2alpha}+frac{2Sinfrac{3alpha-alpha}{2}Cosfrac{3alpha+alpha}{2}}{2Cosfrac{3alpha+alpha}{2}Cosfrac{3alpha-alpha}{2}} =frac{Cos2alpha}{Sin2alpha} +frac{Sinalpha Cos2alpha}{Cos2alpha Cosalpha}=frac{Cos2alpha}{Sin2alpha}+frac{Sinalpha}{Cosalpha}=frac{Cos2alpha Cosalpha+Sin2alpha Sinalpha}{Sin2alpha Cosalpha}=$

$=frac{Cos(2alpha-alpha)}{Sin2alpha Cosalpha}=frac{Cosalpha}{Sin2alpha Cosalpha} =frac{1}{Sin2alpha}\\frac{1}{Sin2alpha}=frac{1}{Sin2alpha }$

Тождество доказано