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

$sqrt{2}-1=frac{1}{sqrt{2}+1 } Rightarrow\\(frac{1}{sqrt{2}+1} )^{x}+(sqrt{2}+1)^{x} -2=0\\(sqrt{2}+1)^{x}=m\\frac{1}{m} +m-2=0\\frac{m^{2}-2m+1}{m}=0\\left { {{m^{2}-2m+1=0 } atop {mneq0 }} right. \\m^{2}-2m+1=0\\(m-1)^{2}=0\\m=1\\(sqrt{2}+1)^{x}=1\\(sqrt{2}+1)^{x} =(sqrt{2}+1)^{0}\\x=0\\Otvet:boxed{0}$

Замена происходит так :

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