Question

Mă poate ajuta cineva ?Va rog ! E urgent ! Multumesc !!

Answer 1

$\it x^2-3x+2=0 \Rightarrow x^2-x-2x+2=0 \Rightarrow x(x-1)-2(x-1)=0 \Rightarrow \\ \\ \Rightarrow (x-1)(x-2)=0 \Rightarrow \begin{cases}\it x-1=0 \Rightarrow x_1=1\\ \\ \it x-2=0 \Rightarrow x_2=2\end{ases}$

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$\it 2x^2-3x=0 \Rightarrow x(2x-3)=0 \Rightarrow \begin{cases}\it x_1=0\\ \\ \it 2x-3=0 \Rightarrow 2x=3 \Rightarrow x_2=\dfrac{3}{2}=1,5\end{cases}$