[tex]m^2x+1=m(x+1) \Rightarrow m^2x+1=mx+m \Rightarrow m^2x+1 -mx-m=0
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\Rightarrow mx(m-1)-(m-1)=0 \Rightarrow mx(m-1) = (m-1)
\\\;\\\Rightarrow x=\dfrac{m-1}{m(m-1)} \Rightarrow x = \dfrac{1}{m} \ solutie \ unica\ \ \forall m\in \mathbb{R}-\{0, 1\}.
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Daca\ m = 0,\ ecuatia\ devine:
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0x+1=0 \Rightarrow 0x = -1 \Rightarrow\ ecuatie\ imposibila \Rightarrow S = \phi
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Daca\ \ m =1, \ ecuatia\ devine:
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x +1 =x+1 \Rightarrow\ x-x = 1-1 \Rightarrow\ 0x=0,\ \ ecuatie\ nedeterminata\Rightarrow S=\mathbb{R}[/tex]
Daca ecuatia admite solutia x = 1/2, vom avea:
[tex]m^2\cdot\dfrac{1}{2}+1 =m\left(\dfrac{1}{2} + 1\right) \Rightarrow m^2\cdot\dfrac{1}{2} +1=m\cdot\dfrac{3}{2} \Rightarrow m^2+2=3m \Rightarrow
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\Rightarrow m^2-3m+2=0 \Rightarrow m_1=1,\ \ m_2=2[/tex]
