Question

Nu știu de câte ori sa mai pun ex ca sa mi se adauge un răspuns
Va implor din tot sufletul meu
Va rog mult
Orice sa știe cum se rezolva sa atașeze un răspuns
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Answer 1

Salut!

Exercitiul 16:

$\begin{aligned}\text{a)} \ x&=\frac23(4\sqrt2-\frac5{\sqrt2})\\x&=\frac23(4\sqrt2-\frac{5\sqrt2}{2})\\x&=\frac23 \cdot \frac{3\sqrt2}{2}\\x&=\sqrt2\end{aligned}$

$\begin{aligned}y&=(\sqrt6+\frac{\sqrt2}{\sqrt3})\div\frac2{\sqrt3}\\y&=\frac{\sqrt{18}+\sqrt2}{\sqrt3}\cdot\frac{\sqrt3}2\\y&=(3\sqrt2+\sqrt2)\frac12\\y&=\frac{4\sqrt2}2\\y&=2\sqrt2\end{aligned}$

$\begin{aligned}\Rightarrow \text{Media geometrica} &= \sqrt{xy}\\&=\sqrt{\sqrt2\cdot2\sqrt2}\\&=\sqrt{2\cdot2}\\&=2\end{aligned}$

$\begin{aligned}\text{b)}\ x&=\frac32(2\sqrt3-\frac4{\sqrt3})\\x&=\frac32(2\sqrt3-\frac{4\sqrt3}3)\\x&=\frac32\cdot\frac{2\sqrt3}3\\x&=\frac{2\sqrt3}2\\x&=\sqrt3\end{aligned}$

$\begin{aligned}y&=(\sqrt6+\frac{\sqrt3}{\sqrt2})\div\frac1{\sqrt2}\\y&=\frac{\sqrt{12}+\sqrt3}{\sqrt2}\cdot\sqrt2\\y&=2\sqrt3+\sqrt3\\y&=3\sqrt3\end{aligned}$

$\begin{aligned}\Rightarrow \text{Media geometrica}=&\sqrt{xy}\\=&\sqrt{\sqrt3\cdot3\sqrt3}\\=&\sqrt{3\cdot3}\\=&3\end{aligned}$

Exercitiul 17:

$\begin{aligned}\text{a)}\ x&=(\sqrt3+(3\sqrt3)^3)\div7\\x&=(\sqrt3+27\cdot3\sqrt3)\div7\\x&=(\sqrt3+81\sqrt3)\div7\\x&=82\sqrt3\div7\\x&=\frac{82\sqrt3}7\end{aligned}$

$\begin{aligned}y&=((\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{\sqrt3})\div\frac1{\sqrt2})^3\\y&=((\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{\sqrt3})\sqrt2)^{-1}\\y&=(\sqrt{3}-\frac{2}{\sqrt3})^3\\y&=(\frac{\sqrt3}3)^3\\y&=\frac{3\sqrt3}27\\y&=\frac{\sqrt3}9\end{aligned}$

$\begin{aligned}\Rightarrow \text{Media geometrica}=&\sqrt{xy}\\=&\sqrt{\frac{82\sqrt3}7\cdot\frac{\sqrt3}9\\=&\sqrt{\frac{246}{63}}\\=&\sqrt\frac{82}{21}\\=&\frac{\sqrt{1722}}{21}\end{aligned}$

$\begin{aligned}\text{b)}\ x&=(\sqrt2+(2\sqrt2)^3)\div3\\x&=(\sqrt2+16\sqrt2)\div3\\x&=17\sqrt2\div3\\x&=\frac{17\sqrt2}3\end{aligned}$

$\begin{aligned}y&=((\frac{\sqrt5}{\sqrt2}-\frac{\sqrt2}{\sqrt5})\div\frac3{\sqrt5})^3\\y&=((\frac{\sqrt5}{\sqrt2}-\frac{\sqrt2}{\sqrt5}\frac{\sqrt5}3)^3\\y&=(\frac{\frac5{\sqrt2}-\sqrt2}{3})^3\\y&=(\frac{\frac3{\sqrt2}}3)^3\\y&=(\frac1{\sqrt2})^3\\y&=(\frac{\sqrt2}2)^3\\y&=\frac{2\sqrt2}8\\y&=\frac{\sqrt2}4\end{aligned}$

$\begin{aligned}\Rightarrow \text{Media geometrica}=& \sqrt{xy}\\=& \sqrt{\frac{17\sqrt2}3\cdot\frac{\sqrt2}4}\\=& \sqrt\frac{17\sqrt2\sqrt2}{12}\\=& \sqrt\frac{34}{12}\\=& \sqrt\frac{17}6\\=& \frac{\sqrt{102}}6\end{aligned}$

-Luke48