Odpowiedź:
Najpierw sprawdzimy m=2
2*(2*2-3)x+5*2-6>0
2x+4>0 , nie jest prawdziwe dla każdego x∈R
m≠2
[tex]\displaystyle (m-2)x^{2} +2(2m-3)x+5m-6 > 0\\\displaystyle \left\{\begin{array}{lrl}\Delta < 0\\a=m-2 > 0\end{array} \right.\\\Delta=4(2m-3)^2-4(m-2)(5m-6)\\\Delta=4(4m^{2} -12m+9)-20m^2+64m-48\\\Delta=16m^{2} -48m+36-20m^{2} +64m-48\\\Delta=-4m+16m-12\\-4m^2+16m-12 < 0/:(-4)\\m^{2} -4m+3 > 0\\m^{2} -3m-m+3 > 0\\m(m-3)-(m-3) > 0\\(m-3)(m-1) > 0\\m\in(-\infty,1)\cup(3,+\infty)\\m-2 > \quad \Rightarrow m > 2\\m\in(-\infty,1)\cup(3,+\infty)\land\quad m\in(2,+\infty)\quad \Rightarrow \underline {m\in(3,+\infty)}[/tex]
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Verified answer
Odpowiedź:
Najpierw sprawdzimy m=2
2*(2*2-3)x+5*2-6>0
2x+4>0 , nie jest prawdziwe dla każdego x∈R
m≠2
[tex]\displaystyle (m-2)x^{2} +2(2m-3)x+5m-6 > 0\\\displaystyle \left\{\begin{array}{lrl}\Delta < 0\\a=m-2 > 0\end{array} \right.\\\Delta=4(2m-3)^2-4(m-2)(5m-6)\\\Delta=4(4m^{2} -12m+9)-20m^2+64m-48\\\Delta=16m^{2} -48m+36-20m^{2} +64m-48\\\Delta=-4m+16m-12\\-4m^2+16m-12 < 0/:(-4)\\m^{2} -4m+3 > 0\\m^{2} -3m-m+3 > 0\\m(m-3)-(m-3) > 0\\(m-3)(m-1) > 0\\m\in(-\infty,1)\cup(3,+\infty)\\m-2 > \quad \Rightarrow m > 2\\m\in(-\infty,1)\cup(3,+\infty)\land\quad m\in(2,+\infty)\quad \Rightarrow \underline {m\in(3,+\infty)}[/tex]