🎬 CLIPURI

\[ \textbf{1.} \text{Dacă } x_1,x_2 \text{ sunt soluțiile reale ale ecuației } x^2+3x+1=0, \text{atunci valoarea expresiei } \left(\frac{x_1}{x_2+1}\right)^2+ \left(\frac{x_2}{x_1+1}\right)^2 \text{ este ?} \]
Rezolvare

\[ \textbf{2. }\; \text{Fie } a,b,c,d\in\mathbb{N}^* \text{ astfel încât } \log_a b=\frac{4}{3} \text{ și } \log_c d=\frac{5}{6}. \text{Dacă } c-a=37,\text{ atunci determinați } b-d. \]
\[ \begin{aligned} \textbf{Rezolvare.}\\ \log_a b=\frac{4}{3} &\Rightarrow a^{\frac{4}{3}}=b \Rightarrow a^4=b^3 \Rightarrow a \text{ cub perfect}.\\ \text{Fie } a=x^3 &\Rightarrow (x^3)^4=b^3 \Rightarrow x^{12}=b^3 \Rightarrow b=x^4.\\ \log_c d=\frac{5}{6} &\Rightarrow c^{\frac{5}{6}}=d \Rightarrow c^5=d^6.\\ \text{Fie } c=y^6 &\Rightarrow (y^6)^5=d^6 \Rightarrow y^{30}=d^6 \Rightarrow d=y^5.\\ c-a=37 &\Rightarrow y^6-x^3=37,\quad x,y\in\mathbb{N}^*.\\ &2^6-3^3=64-27=37 \Rightarrow y=2,\; x=3.\\ &b=3^4=81,\qquad d=2^5=32.\\ &b-d=81-32=49. \end{aligned} \]
3. Fie sistemul de ecuații \[ \left\{ \begin{aligned} 2mx+y+(m+1)z&=2m+1\\ (m+2)x+(m+1)y+(m+2)z&=2\\ 3mx+y+(2m+1)z&=1 \end{aligned} \right. \] \(m\in\mathbb{R}\). Determinați \(m\in\mathbb{R}\) pentru care sistemul este incompatibil.
\[ \begin{gathered} \textbf{Rezolvare} \\[8pt] A= \begin{pmatrix} 2m&1&m+1\\ m+2&m+1&m+2\\ 3m&1&2m+1 \end{pmatrix} \\[10pt] \text{sistem incompatibil }\Leftrightarrow \begin{cases} \det A=0\\ \Delta_c\neq0 \end{cases} \\[12pt] \det A= \begin{vmatrix} 2m&1&m+1\\ m+2&m+1&m+2\\ 3m&1&2m+1 \end{vmatrix} = \begin{vmatrix} m-1&1&m+1\\ 0&m+1&m+2\\ m-1&1&2m \end{vmatrix} \\[12pt] =(m-1) \begin{vmatrix} 0&m+1&m\\ 1&m+1&1\\ 1&1&2m \end{vmatrix} = (m-1) \begin{vmatrix} 0&0&-m\\ 1&m+1&1\\ 1&1&2m \end{vmatrix} \\[12pt] =(m-1)m \begin{vmatrix} 1&m+1\\ 1&1 \end{vmatrix} =(m-1)m(m+1) \\[12pt] \det A=0 \Rightarrow m\in\{-1,0,1\} \\[14pt] d_p= \begin{vmatrix} 2m&1\\ m+2&m+1 \end{vmatrix} =2m(m+1)-(m+2) =2m^2+2m-m-2 =2m^2+m-2 \\[10pt] d_p\neq0 \text{ pentru } m\in\{-1,0,1\} \\[12pt] r=2,\qquad s=3 \Rightarrow s-r=1 \\[12pt] \Delta_c= \begin{vmatrix} 2m&1&2m+1\\ m+2&m+1&2\\ 3m&1&1 \end{vmatrix} = \begin{vmatrix} 2m&1&2m\\ m+2&m+1&1-m\\ 3m&1&0 \end{vmatrix} \\[12pt] = \begin{vmatrix} 0&1&2m\\ 2m+1&m+1&1-m\\ 3m&1&0 \end{vmatrix} \\[12pt] = -\begin{vmatrix} 2m+1&1-m\\ 3m&0 \end{vmatrix} +2m \begin{vmatrix} 2m+1&m+1\\ 3m&1 \end{vmatrix} \\[12pt] =3m(1-m)+2m(2m+1-3m^2) \\[8pt] =3m-3m^2+2m(-3m^2-m+1) \\[8pt] =-6m^3-3m^2+m \\[8pt] =m(-6m^2-3m+1) \\[12pt] \Delta_c\neq0 \Rightarrow m\neq0 \\[10pt] \boxed{m\in\{-1,1\}} \end{gathered} \]