Question

Cine ma poate ajuta la ex 57?

Answer 1

[tex]57 d) \ \ \sqrt{2x^2+x+1}=2x^2+x-1 \\ Notam:\ 2x^2+x+1 =t, \ t\ \textgreater \ 0 \\ 2x^2+x-1=2x^2+x+1-2=t-2,\ t-2\ \textgreater \ 0\ \Rightarrow t\ \textgreater \ 2 \\ Ecuatia\ devine: \\ \sqrt{t} =t-2 \Rightarrow (\sqrt{t})^2 =(t-2)^2 \Rightarrow t=t^2-4t+4 \Rightarrow t^2-5t+4=0[/tex]

[tex]\Rightarrow t_1=1,\ nu\ convine\ (pentru\ ca\ t\ \textgreater \ 2) \\ t_2=4 \\ Revenim\ asupra\ notatiei: \\ t=4 \Rightarrow 2x^2+x+1=4 \Rightarrow 2x^2+x-3=0 \\ Obtinem: \\ \it x_1=-\dfrac{3}{2},\ \ x_2=1. [/tex]

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