Salut!
Exercitiul 16:
[tex]\begin{aligned}\text{a)} \ x&=\frac23(4\sqrt2-\frac5{\sqrt2})\\x&=\frac23(4\sqrt2-\frac{5\sqrt2}{2})\\x&=\frac23 \cdot \frac{3\sqrt2}{2}\\x&=\sqrt2\end{aligned}[/tex]
[tex]\begin{aligned}y&=(\sqrt6+\frac{\sqrt2}{\sqrt3})\div\frac2{\sqrt3}\\y&=\frac{\sqrt{18}+\sqrt2}{\sqrt3}\cdot\frac{\sqrt3}2\\y&=(3\sqrt2+\sqrt2)\frac12\\y&=\frac{4\sqrt2}2\\y&=2\sqrt2\end{aligned}[/tex]
[tex]\begin{aligned}\Rightarrow \text{Media geometrica} &= \sqrt{xy}\\&=\sqrt{\sqrt2\cdot2\sqrt2}\\&=\sqrt{2\cdot2}\\&=2\end{aligned}[/tex]
[tex]\begin{aligned}\text{b)}\ x&=\frac32(2\sqrt3-\frac4{\sqrt3})\\x&=\frac32(2\sqrt3-\frac{4\sqrt3}3)\\x&=\frac32\cdot\frac{2\sqrt3}3\\x&=\frac{2\sqrt3}2\\x&=\sqrt3\end{aligned}[/tex]
[tex]\begin{aligned}y&=(\sqrt6+\frac{\sqrt3}{\sqrt2})\div\frac1{\sqrt2}\\y&=\frac{\sqrt{12}+\sqrt3}{\sqrt2}\cdot\sqrt2\\y&=2\sqrt3+\sqrt3\\y&=3\sqrt3\end{aligned}[/tex]
[tex]\begin{aligned}\Rightarrow \text{Media geometrica}=&\sqrt{xy}\\=&\sqrt{\sqrt3\cdot3\sqrt3}\\=&\sqrt{3\cdot3}\\=&3\end{aligned}[/tex]
Exercitiul 17:
[tex]\begin{aligned}\text{a)}\ x&=(\sqrt3+(3\sqrt3)^3)\div7\\x&=(\sqrt3+27\cdot3\sqrt3)\div7\\x&=(\sqrt3+81\sqrt3)\div7\\x&=82\sqrt3\div7\\x&=\frac{82\sqrt3}7\end{aligned}[/tex]
[tex]\begin{aligned}y&=((\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{\sqrt3})\div\frac1{\sqrt2})^3\\y&=((\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{\sqrt3})\sqrt2)^{-1}\\y&=(\sqrt{3}-\frac{2}{\sqrt3})^3\\y&=(\frac{\sqrt3}3)^3\\y&=\frac{3\sqrt3}27\\y&=\frac{\sqrt3}9\end{aligned}[/tex]
[tex]\begin{aligned}\Rightarrow \text{Media geometrica}=&\sqrt{xy}\\=&\sqrt{\frac{82\sqrt3}7\cdot\frac{\sqrt3}9\\=&\sqrt{\frac{246}{63}}\\=&\sqrt\frac{82}{21}\\=&\frac{\sqrt{1722}}{21}\end{aligned}[/tex]
[tex]\begin{aligned}\text{b)}\ x&=(\sqrt2+(2\sqrt2)^3)\div3\\x&=(\sqrt2+16\sqrt2)\div3\\x&=17\sqrt2\div3\\x&=\frac{17\sqrt2}3\end{aligned}[/tex]
[tex]\begin{aligned}y&=((\frac{\sqrt5}{\sqrt2}-\frac{\sqrt2}{\sqrt5})\div\frac3{\sqrt5})^3\\y&=((\frac{\sqrt5}{\sqrt2}-\frac{\sqrt2}{\sqrt5}\frac{\sqrt5}3)^3\\y&=(\frac{\frac5{\sqrt2}-\sqrt2}{3})^3\\y&=(\frac{\frac3{\sqrt2}}3)^3\\y&=(\frac1{\sqrt2})^3\\y&=(\frac{\sqrt2}2)^3\\y&=\frac{2\sqrt2}8\\y&=\frac{\sqrt2}4\end{aligned}[/tex]
[tex]\begin{aligned}\Rightarrow \text{Media geometrica}=& \sqrt{xy}\\=& \sqrt{\frac{17\sqrt2}3\cdot\frac{\sqrt2}4}\\=& \sqrt\frac{17\sqrt2\sqrt2}{12}\\=& \sqrt\frac{34}{12}\\=& \sqrt\frac{17}6\\=& \frac{\sqrt{102}}6\end{aligned}[/tex]
-Luke48
