Question

Розв'яжіть рівняння
(3 - 2√2)^х + (3 + 2√2)^х = 6​

Answer 1

Объяснение:

$(3-2\sqrt{2} )^x+(3+2\sqrt{2})^x=6.$

Пусть (3-2√2)ˣ=t,  (3+2√3)ˣ=v.          ⇒

t+v=6

t*v=(3-2√2)*(3+2√2)=3²-(2√2)²=9-8=1.       ⇒

$\left \{ {{t+v=6} \atop {t*v=1}} \right. \ \ \ \ \left \{ {{v=6-t} \atop {t*(6-t)=1}} \right.\ \ \ \ \left \{ {{y=6-t} \atop {6t-t^2=1}} \right.\ \ \ \ \left \{ {{y=6-t} \atop {t^2-6t+1=0}} \right. \ \ \ \ \left \{ {{y=6-t} \atop {D=32\ \ \sqrt{D}=4\sqrt{2} } } \right.\ \ \ \ \left \{ {{v_1=3+2\sqrt{2}\ \ v_2=3-2\sqrt{2} } \atop {t_1=3-2\sqrt{2}\ \ t_2=3+2\sqrt{2} }}. \right.$$t_1=(3-2\sqrt{2)^x}=3-2\sqrt{2}\ \ \ \ \Rightarrow\ \ \ \ x_1=1.\\v_1=(3+2\sqrt{2})^x=3+2\sqrt{2} \ \ \ \ \Rightarrow\ \ \ \ x_2=1.\\t_2=(3-2\sqrt{2})^x= 3+2\sqrt{2} \ \ \ \ \Rightarrow\ \ \ \ x_3=-1 .\\v_2=(3+2\sqrt{2})^x= 3-2\sqrt{2} \ \ \ \ \Rightarrow\ \ \ \ x_4=-1.$

ответ: x₁=1,  x₂=-1.