Explicație pas cu pas:
[tex](m - 2) {x}^{2} - 2mx + 2m - 3 = 0 \\ {x}^{2} - \frac{2m}{m - 2} \cdot x + \frac{2m - 3}{m - 2} = 0; \ \ m \not = 2[/tex]
din Relațiile lui Viete:
[tex]\boxed {\red {\bf {x}^{2} - Sx + P = 0}}[/tex]
[tex]\red { S = x_{1} + x_{2}}; \ \ \ \red {P = x_{1}x_{2}}[/tex]
[tex]S = \frac{2m}{m - 2}; \ \ \ P = \frac{2m - 3}{m - 2} \\ [/tex]
[tex]\frac{1}{ {x_{1}}^{2} } + \frac{1}{{x_{2}}^{2} } = \frac{x_{1}^{2} + x_{2}^{2}}{x_{1}^{2}x_{2}^{2}} = \frac{(x_{1} + x_{2})^{2} - 2x_{1}x_{2}}{(x_{1}x_{2})^{2}} = \frac{S^{2} - 2P}{P^{2}} \\ \implies \bf \frac{S^{2} - 2P}{P^{2}} = 2 \iff S^{2} - 2P = 2P^{2}[/tex]
[tex]\Big(\frac{2m}{m - 2}\Big)^{2} - 2\Big(\frac{2m - 3}{m - 2}\Big) = 2\Big(\frac{2m - 3}{m - 2}\Big)^{2}\\[/tex]
[tex]4 {m}^{2} - 2(2m - 3)(m - 2) = 2 {(2m - 3)}^{2}[/tex]
[tex]4 {m}^{2} - 19m + 15 = 0 \\ (4m - 15)(m - 1) = 0[/tex]
[tex]4m - 15 = 0 \implies \bf m = \frac{15}{4} \\ m - 1 = 0 \implies \bf m = 1[/tex]
