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Piotrek89 pisze:\(\displaystyle{ \frac{2x-1}{x-1} - \frac{x+1}{2x+1}+m=0}\)

Przekształcam równanie:

\(\displaystyle{ \frac{(4x^{2}-1)-(x^{2}-1)}{(x-1)(2x+1)} +m=0}\)
\(\displaystyle{ \frac{3x^{2}+m(x-1)(2x+1)}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{3x^{2}+(mx-m)(2x+1)}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{3x^{2}+2mx^{2}+mx-2mx-m}{(x-1)(2x+1)}=0}\)
\(\displaystyle{ \frac{(3+2m)x^{2}-mx-m}{(x-1)(2x+1)}=0}\)

\(\displaystyle{ (3+2m)x^{2}-mx-m=0 \wedge (x-1)(2x+1)\neq 0}\)

\(\displaystyle{ D_{f}=R \backslash \lbrace 1,-\frac{1}{2} \rbrace}\)

\(\displaystyle{ (3+2m)x^{2}-mx-m=0}\)
Założenia:

\(\displaystyle{ \begin{cases} \Delta>0\\x_{1}x_{2}0}\)
\(\displaystyle{ m(9m+12)>0}\)
\(\displaystyle{ m0}\)

2 warunek:
\(\displaystyle{ x_{1}x_{2}}\)