[tex]\displaystyle 188.~ Observam~ca~\sqrt{5+2\sqrt{6}} \cdot \sqrt{5-2 \sqrt{6}}= \sqrt{25-24}=1. \\ \\ Astfel,~notand~y= \left(\sqrt{5+2 \sqrt{6}}\right)^x,~vom~avea~ \left(\sqrt{5-2 \sqrt{6}}\right)^x= \frac{1}{y}. \\ \\ Ecuatia~devine:~y+ \frac{1}{y}=10 \Leftrightarrow~y^2-10y+1=0. \\ \\ Obtinem~y_{1,2}= \frac{10 \pm 4 \sqrt{6}}{2}=5 \pm 2 \sqrt{6}.[/tex]
[tex]\displaystyle Avem~doua~posibilitati: \\ \\ i) \left(\sqrt{5+2 \sqrt{6}}\right)^x= 5+2 \sqrt{6} \Rightarrow x=2. \\ \\ ii)~ \left(\sqrt{5+2 \sqrt{6}}\right)^x=5- 2\sqrt{x} \Rightarrow x=-2. \\ \\ Raspuns:~b).[/tex]
[tex]\displaystyle 189.~Notam~|x|=t \ge 0~si~consideram~functia~f: [0;+ \infty)\rightarrow \mathbb{R}, \\ \\ f(t)=3^t+4^t+5^t. \\ \\ Observam~ca~functia~f~este~suma~de~trei~functii~strict~crescatoare. \\ \\ Deci~f~este~injectiva,~si~deci,~ecuatii~f(t)=12~admite~cel~mult ~o\\ \\ solutie.~Observam~ca~f(1)=12 \Rightarrow t=1. \\ \\ Deci~|x|=1 \Rightarrow x=1~sau~x=-1. \\ \\ Raspuns:~b).[/tex]
[tex]\displaystyle 190.~Domeniul~de~d {} efinitie:~x\ \textgreater \ 0. \\ \\ Notam~v=lgx.~Atunci~x=10^v,~iar~ecuatia~devine: \\ \\ 12v^2-v-1=0. \\ \\ \Delta=49. \\ \\ v_{1,2}= \frac{1 \pm 7}{24}= \left \{ {{\frac{1}{3}} \atop {-\frac{1}{4}}} \right. . \\ \\ Astfel,~x \in \left\{ 10^ \frac{1}{3}; 10^{-\frac{1}{4}}\right\}= \left \{ \sqrt[3]{10} ; \sqrt[4]{10^{-1}} \right\}. \\ \\ Raspuns:~c).[/tex]
