Răspuns:
[tex]D. \ \frac{15}{4} \ si \ 1[/tex]
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[/tex]
din Relațiile lui Viete:
[tex]S = x_{1} + x_{2} = \frac{2m}{m - 2} \\ [/tex]
[tex]P = x_{1} x_{2} = \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} \cdot {x_{2}}^{2}} = \frac{{x_{1}}^{2} + {x_{2}}^{2}}{{x_{1}}^{2} \cdot {x_{2}}^{2}} = \\ = \frac{{x_{1}}^{2} + {x_{2}}^{2} + 2x_{1} \cdot x_{2} - 2x_{1} \cdot x_{2}}{{x_{1}}^{2} \cdot {x_{2}}^{2}} = \\ = \frac{ {(x_{1} + x_{2})}^{2} - 2 (x_{1}x_{2}) }{(x_{1}x_{2})^{2}} = \frac{ {S}^{2} - 2P }{ {P}^{2} }[/tex]
[tex]\frac{{S}^{2}-2P }{ {P}^{2} } = 2 \iff {S}^{2} - 2P = 2{P}^{2} \\ {S}^{2} - 2P - 2{P}^{2} = 0[/tex]
[tex]\Big(\frac{2m}{m - 2}\Big)^{2}-2\cdot\frac{2m - 3}{m - 2}-2\cdot\Big(\frac{2m - 3}{m - 2}\Big)^{2} = 0 \\[/tex]
[tex]4m^{2} - 2(2m - 3)(m - 2) - 2(2m - 3)^{2} = 0[/tex]
[tex]4m^{2} - 4m^{2} + 14m - 12 - 8m^{2} + 24m - 18 = 0[/tex]
[tex]-8m^{2} + 38m - 30 = 0 \iff 4m^{2} - 19m + 15 = 0[/tex]
[tex]4m^{2} - 4m - 15 m + 15 = 0 \iff 4m(m - 1) - 15(m - 1) = 0[/tex]
[tex](m - 1)(4m - 15) = 0[/tex]
[tex]m - 1 = 0 \implies m = 1[/tex]
[tex]4m - 15 = 0 \implies m = \frac{15}{4}\\[/tex]
