Problemă rezolvată dintre subiectele propuse pentru clasa a X-a la ONGM 2020-2021.
Enunț
\text{Dacă } a,b >0, \ a^2 + b^2 = 11 \cdot ab \ \ \\\;\\ \text{ și} \ \dfrac{lg a + lg b }{2} = lg \dfrac{a + b }{p} ,\qquad\qquad\qquad \\ \text{ atunci numărul p este egal cu: \qquad\qquad\qquad\qquad\qquad\qquad\qquad} \\\;\\ \ \ \
A. \ 3 \qquad\qquad B. \ 7 \qquad\qquad \\\;\\ C. \ \sqrt[]{7} \quad\qquad D. \ \sqrt[]{11}\qquad\qquad E. \ \sqrt[]{13}Rezolvare
\newline
\left .
\begin{array}{ll}
a, b > 0 \\ \; \\
\dfrac{a + b }{p} > 0
\end{array}
\right \} \implies p > 0
\\ \;\\
\dfrac{lg a + lg b }{2} = lg \dfrac{a + b }{p} \Leftrightarrow \\ \; \\ \Leftrightarrow lg a + lg b = 2 \cdot lg \dfrac{a + b }{p} \Leftrightarrow \\ \; \\ \Leftrightarrow lg a + lg b = lg\bigg(\dfrac{a + b }{p} \bigg)^2 \Leftrightarrow \\ \; \\ \Leftrightarrow lg (a \cdot b) = lg\bigg(\dfrac{a + b }{p} \bigg)^2 \Leftrightarrow \\\ \; \\ \Leftrightarrow a \cdot b = \bigg(\dfrac{a + b }{p} \bigg)^2 \Leftrightarrow \\\ \; \\
\Leftrightarrow a \cdot b = \dfrac{(a + b)^2 }{p^2} \\ \;\\\quad a^2 + b^2 = 11 ab \ \Leftrightarrow \\\ \; \\ \Leftrightarrow a^2 + 2ab + b^2 = 13ab \Leftrightarrow \\\ \; \\ \Leftrightarrow (a + b)^2 = 13 ab
\left .
\begin{array}{ll}
a \cdot b = \dfrac{(a + b)^2 }{p^2} \\ \;\\
(a + b)^2 = 13 ab
\end{array}
\right \} \implies \\\ \; \\ \implies \cancel{a \cdot b} = \dfrac{ 13 \cdot \cancel{ab} }{p^2} \Leftrightarrow \\\ \; \\ \Leftrightarrow p^2 = 13 \Leftrightarrow \\\ \; \\
\left .
\begin{array}{ll}
\Leftrightarrow p = |\sqrt[]{13} | \\ \; \\ \quad p > 0
\end{array}
\right \} \implies p = \sqrt[]{13} \text{Răspunsul este } \textbf{ E}, p = \sqrt[]{13} .
