0,(1) = 1 / 9
Vom folosi aceste formule:
[tex]a^ \frac{p}{q}= \sqrt[q]{a^p} [/tex]
[tex]log_{a^q}(x^p)= \frac{p}{q}log_a(x) [/tex]
1.
[tex]\sqrt[6]{64}+512^{0,(1)}-log_{ \sqrt{3} } \frac{1}{9} = \sqrt[6]{2^6}+512^{ \frac{1}{9} }-log_{3^{ \frac{1}{3} }}( \frac{1}{3^2} )=\\ =2+ \sqrt[9]{2^9}-3log_3(3^{-2}) =4-3\cdot(-2)=10 [/tex]
2.
[tex] \sqrt{2}^x\cdot \sqrt[3]{2}^{x+1}=0,25^{-1}\\\\ 2^{ \frac{x}{2} }\cdot2^{ \frac{x+1}{3} }= (\frac{1}{4} )^{-1}\\\\ 2^{ \frac{x(x+1)}{6} }=2^2\\ \frac{x(x+1)}{6}=2\\ x(x+1)=12\\ x^2+x-12=0\\ x_1=3\\ x_2=-4 [/tex]
3.
Aici se vede ca 3 - 2√2 este subunitar, pozitiv, deci (3 - 2√2)^x este subunitar ==> 33 < (3 + 2√2)^x > 34
Daca x este intreg, atunci x = 2. Nu poate fi mai mic sau mai mare deoarece (3 + 2√2)^x nu se va incadra intre 33 si 34:
(3 + 2√2)² + (3-2√2)² = 9 + 12√2 + 8 + 9 - 12√2 + 8 = 34
