Question

Помогите решить уравнение сроооочно!

Answer 1

(√2-1)^x+(√2+1)^x -2=0

заменим √2-1=1/(√2+1)

t=(√2-1)^x >0

t+1/t-2=0

t²-2t+1=0

(t-1)²=0

t=1

(√2-1)^x=1

(√2-1)^x=(√2-1)^0

x=0

Answer 2

$\sqrt{2}-1=\frac{1}{\sqrt{2}+1 } \Rightarrow\\\\(\frac{1}{\sqrt{2}+1} )^{x}+(\sqrt{2}+1)^{x} -2=0\\\\(\sqrt{2}+1)^{x}=m\\\\\frac{1}{m} +m-2=0\\\\\frac{m^{2}-2m+1}{m}=0\\\\\left \{ {{m^{2}-2m+1=0 } \atop {m\neq0 }} \right. \\\\m^{2}-2m+1=0\\\\(m-1)^{2}=0\\\\m=1\\\\(\sqrt{2}+1)^{x}=1\\\\(\sqrt{2}+1)^{x} =(\sqrt{2}+1)^{0}\\\\x=0\\\\Otvet:\boxed{0}$

Замена происходит так :

$\frac{1}{\sqrt{2}+1}=\frac{1*(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2-1)}}=\frac{\sqrt{2}-1}{(\sqrt{2})^{2}-1^{2}} =\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\\\\\frac{1}{\sqrt{2}+1 }=\sqrt{2}-1$