Question

у системі рівнянь {х+у-6ху=-34;ху(х+у)=56
хееееелп​

Answer 1

Объяснение:

$\displaystyle\\\left \{ {{x+y-6xy=-34} \atop {xy*(x+y)=56}} \right. .$

Пусть х+у=t,   xy=v.         ⇒

$\displaystyle\\\left \{ {{t-6v=-34} \atop {tv=56}} \right. \ \ \ \ \left \{ {{t=6v-34} \atop {(6v-34)*v=56}} \right. \ \ \ \ \left \{ {{t=6v-34} \atop {6v^2-34v-56=0}} \right. \\\\\\\left \{ {{t=6v-34} \atop {6v^2-42v+8v-56=0}} \right.\ \ \ \ \left \{ {{t=6v-34} \atop {6v*(v-7)+8*(v-7)=0\\}} \right. \\\\\\\left \{ {{t=6v-34} \atop {(v-7)*(6v+8)=0}} \right. \ \ \ \ \left \{ {{t_1=8\ \ \ \ t_2=-42} \atop {v_1=7\ \ \ \ v_2=-\frac{4}{3} }} \right. .$

$\displaystyle\\1.\\\\\left \{ {{x+y=8} \atop {xy=7}} \right.\ \ \ \ \left \{ {{y=8-x} \atop {x*(8-x)=7}} \right. \ \ \ \ \left \{ {{y=8-x} \atop {8x-x^2=7}} \right.\ \ \ \ \left \{ {{y=8-x} \atop {x^2-8x+7=0}} \right. \\\\\\\left \{ {{y=8-x} \atop {x^2-7x-x+7=0}} \right.\ \ \ \ \left \{ {{y=8-x} \atop {x*(x-7)-(x-7)=0}} \right. \ \ \ \ \left \{ {{y=8-x} \atop {(x-7)*(x-1)=0}} \right. \\\\\\\left \{ {{y_1=1\ \ \ y_2=7 \atop {x_1=7\ \ \ \ x_2=1}} \right. .$

$\displaystyle\\2.\\\\\left \{ {{x+y=-42} \atop {xy=-\frac{4}{3}\ |*3 }} \right. \ \ \ \ \left \{ {{y=-42-x} \atop {3x*(-42-x)=-4\ |*(-1)}} \right.\ \ \ \ \left \{ {{y=-42-x} \atop {126x+3x^2=4}} \right. \\\\3x^2+126x-4=0\\\\D=(-126)^2-4*3*(-4)=15876+48=15924=4*3981\\\\x_3=-21-\frac{\sqrt{3981} }{3} \ \ \ \ x_4=-21+\frac{\sqrt{3981} }{3} \\\\y_3=-21+\frac{\sqrt{3981} }{3} \ \ \ \ y_4=-21-\frac{\sqrt{3981} }{3} .$