Question

решить систему уравнений
9^x -2^y =1
{
9^-x -2^-y=-1/6

Answer 1

Ответ:  x=0,5   y=1.

Объяснениe:

$\left \{ {{9^{x}-2^{y} =1} \atop {\frac{1}{9^{x} } -\frac{1}{2^{y} } =-\frac{1}{6} }} \right.$

Пусть:  

$9^{x}=u>0;2^{y}=v>0.\\\left \{ {{u-v=1} \atop {\frac{1}{u}-\frac{1}{v} } =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\frac{v-u}{u*v} =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\\\frac{u-v}{u*v} =\frac{1}{6} }} ;\right. \;\left \{ {{u=v+1} \atop {\frac{1}{u*v} =\frac{1}{6} }} \right. ;\left \{ {{u=v+1} \atop {u*v=6}} ;\right.\\\left \{ {{u=v+1} \atop {v*(v+1)=6}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v-6 =0}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v+2v-2v-6 =0}} \right. ;$

$\left \{ {{u=v+1} \atop {v^{2} +3v-2v-6=0}} \right. ;\left \{ {{u=v+1} \atop {v*(v+3)-2*(v+3)=0}} \right. ;\left \{ {{u=v+1} \atop {(v+3)*(v-2)=0}} \right. ;\left \{ {{u_{1}=-2 ;u_{2} =3} \atop {v_{1} =-3;v_{2} =2}} \right.$

u₁=-2 ∉;   v₁=-3 ∉    ⇒

$\left \{ {{u=9^{x} =3} \atop {v^{2} =2^{y} =2}} \right. ;\left \{ {{3^{2x} =3^{1} } \atop {2^{y} =2^{1} }} \right. ;\left \{ {2x=1|:2} \atop {y=1}} \right. ;\left \{ {{x=0,5} \atop {y=1}} \right. .$