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Szemek pisze:\(\displaystyle{ \frac{2x-1}{x-1}-\frac{x+1}{2x+1}+m=0}\)
\(\displaystyle{ \frac{(2x-1)(2x+1)-(x+1)(x-1)+m(x-1)(2x+1)}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{4x^2-1-(x^2-1)+m(2x^2+x-2x-1)}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{3x^2+m(2x^2-x-1)}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{3x^2+2{mx^2}-{mx}-{m}}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ D_r=R-\{-\frac{1}{2},1\}}\)
\(\displaystyle{ \frac{3x^2+2{mx^2}-{mx}-{m}}{(x-1)(2x+1)}=0}\)|\(\displaystyle{ \cdot (x-1)(2x+1), x \in D_r}\)
\(\displaystyle{ {(2m+3)x^2}-{mx}-{m}}=0}\)

Żeby równanie miało dwa pierwiastki różnych znaków muszą być spełnione warunki:
\(\displaystyle{ \begin{cases} 2m+3 \neq 0 \\ \Delta_x >0 \\ x_1 x_2 < 0 \end{cases}}\)
\(\displaystyle{ \begin{cases} m \neq -1\frac{1}{2} \\ m^2-4 \cdot (2m+3)\cdot (-m) >0 \\ \frac{-m}{2m+3}< 0 \end{cases}}\)
\(\displaystyle{ \begin{cases} m \neq -1\frac{1}{2} \\ m^2 + 8m^2+ 12m >0 \\ -m\cdot(2m+3)< 0 \end{cases}}\)
\(\displaystyle{ \begin{cases} m \neq -1\frac{1}{2} \\ m(9m+ 12) >0 \\ m(2m+3)> 0 \end{cases}}\)
\(\displaystyle{ \begin{cases} m -1\frac{1}{2} \\ m\in (-\infty , -1\frac{1}{3}) \cup (0,+\infty) \\ m (-\infty , -1\frac{1}{2}) \cup (0,+\infty) \end{cases}}\)
\(\displaystyle{ m (-\infty , -1\frac{1}{2}) \cup (0,+\infty)}\)

Odpowiedź:Równanie \(\displaystyle{ \frac{2x-1}{x-1}-\frac{x+1}{2x+1}+m=0}\) ma dwa pierwiastki różnych znaków dla \(\displaystyle{ m (-\infty , -1\frac{1}{2}) \cup (0,+\infty)}\)