Question

URGENTT!!! Va roooggg!!!
Daauuuu COROOOANA
Rezolvati in Q ecuatia:
a) $\frac{2x-1}{x-1} = \frac{5 x^{2} -1}{x-1}$
b) $\frac{ x^{2} +2}{x+2} = \frac{2x-1}{x+2}$
c) $\frac{x(x-1)}{1-3x} = \frac{ x^{2} +4x}{3x-1}$
d) $\frac{ x^{2} }{ x^{2} +5} = \frac{2x(x-3)}{ x^{2} +5}$

Answer 1

   
[tex]a) \\ \frac{2x-1}{x-1} = \frac{5 x^{2} -1}{x-1} ~~~~~| \cdot (x-1) ~~~~|~Conditie: ~~~ x \neq 1 \\ 2x-1 = 5x^2-1 \\ 5x^2-2x-1+1=0 \\ 5x^2-2x=0 \\ x(5x-2)=0 \\ x_1 = \boxed{0} \\ x_2= \boxed{\frac{2}{5}} [/tex]


b) Punctul b este gresit.
Nu se incadreaza in forma celorlalte.
Daca l-asi rezolva ar rezulta radacini "complex-conjugate" 
si nu cred ca stii ce sunt alea. 


[tex]c) \\ \frac{x(x-1)}{1-3x} = \frac{x^2+4x}{3x-1} \\ \frac{x(1-x)}{3x-1} = \frac{x^2+4x}{3x-1} ~~~~~|\cdot (3x-1)~~~~~|~Conditie: x \neq \frac{1}{3} \\ x(1-x)=x^2+4x \\ x-x^2=x^2+4x \\ x^2+4x + x^2-x =0 \\ 2x^2 +3x =0 \\ x(2x +3)=0 \\ x_1=\boxed{0} \\ x_2 = \boxed{- \frac{3}{2} } [/tex]


[tex]d) \\ \frac{x^2}{x^2+5} = \frac{2x(x-3)}{x^2+5} ~~~~~| \cdot (x^2+5) ~~~~~| Conditie: x \neq i \sqrt{5} \\ x^2= 2x(x-3) \\ x^2= 2x^2-6x \\ 2x^2 - x^2 -6x=0 \\ x^2 -6x=0 \\ x(x -6)=0 \\ x_1 = \boxed{0} \\ x_2 = \boxed{6} [/tex]