Номер 6 !!!!!!!!!!!!!. Created by NoMind04. algebra-ru

Номер 6 !!!!!!!!!!!!!...

NoMind04 @NoMind04

July 2022 1 19 Report

Номер 6 !!!!!!!!!!!!!

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$(\sqrt2-1)^{x}+(\sqrt2+1)^{x}-2=0$

Если $ab=1$ , то $b=\frac{1}{a}$ и выражения a и b называются взаимно обратными. Выражение b обозначается $b=a^{-1}\; ,\; \; a^{-1}=\frac{1}{a}$ .

$(\sqrt2-1)\cdot (\sqrt2+1)=(\sqrt2)^2-1^2=2-1=1\; \; \Rightarrow \\\\\sqrt2-1=\frac{1}{\sqrt2+1} \; \; \Rightarrow \; \; (\sqrt2-1)^{x}= \frac{1}{(\sqrt2+1)^{x}} \\\\t=(\sqrt2+1)^{x}\ \textgreater \ 0\; ,\; \; t^{-1}=\frac{1}{t}=(\sqrt2-1)^{x}\\\\\frac{1}{t}+t-2=0\; ,\; \; \frac{t^2-2t+1}{t} =0\; ,\; \; \frac{(t-1)^2}{t}=0\; ,\\\\(t-1)^2=0\; ,\; \; t\ne 0\\\\t=1\\\\(\sqrt2+1)^{x}=1\\\\(\sqrt2+1)^{x}=(\sqrt2+1)^0\\\\\underline {x=0}$

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