Question

(3-2√2)^x+(3+2√2)^x=6

Answer 1

Ответ:

Объяснение:

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

умножим обе части уравнения на   $\bf(3+2\sqrt{2} )^x\neq 0$

получим

$\displaystyle\bf\bigg((3-2\sqrt{2} )(3+2\sqrt{2} )\bigg)^x+(3+2\sqrt{2} )^{2x}=6(3+2\sqrt{2} )^x\\\\(3+2\sqrt{2} )^{2x}-6(3+2\sqrt{2} )^x+(3^2-4\cdot2)^x=0\\\\(3+2\sqrt{2} )^{2x}-6(3+2\sqrt{2} )^x+1=0\\\\t=(3+2\sqrt{2} )^x\\\\t^2-6t+1=0\\\\D=b^2-4ac=36-4=32\\\\t_1=\frac{6-4\sqrt{2} }{2} =3-2\sqrt{2}\\\\(3+2\sqrt{2} )^x =3-2\sqrt{2} =\frac{(3-2\sqrt{2})(3+2\sqrt{2})}{3+2\sqrt{2}} =(3+2\sqrt{2})^{-1};x_1=-1$

$\displaystyle\bf\\t_2=\frac{6+4\sqrt{2} }{2} =3+2\sqrt{2}\\\\(3+2\sqrt{2} )^x =(3+2\sqrt{2})^1 ;\ \ \ x_2=1\\\\Otvet:\pm1$