Question

Va rog sa ma ajutati cu aceasta ecuatie. Nivel clasa a 9-a. Am tot incercat dar imi da delta un numar foarte mare care nu este patrat perfect​

Answer 1

Răspuns:

$\dfrac{x(2x-3)}{2}+\dfrac{(3x-1)^{2}}{5}=1+\frac{(x+3)^{2}}{5} ,~~\dfrac{x(2x-3)}{2}-1+\dfrac{(3x-1)^{2}}{5}-\frac{(x+3)^{2}}{5},~~\frac{x(2x-3)-2}{2}+\frac{(3x-1)^{2}-(x+3)^{2}}{5}=0,~~\dfrac{2x^{2}-3x-2}{2} +\dfrac{(3x-1+x+3)(3x-1-x-3)}{5}=0,~~ \dfrac{2x^{2}-3x-2}{2} +\dfrac{(4x+2)(2x-4)}{5}=0, ~~\dfrac{2x^{2}-3x-2}{2} +\dfrac{2*(2x+1)*2*(x-2)}{5}=0~|*10,$5(2x²-3x-2)+8(2x²-3x-2)=0, ⇒13·(2x²-3x-2)=0, ⇒2x²-3x-2=0,

Δ=(-3)²-4·2·(-2)=9+16=25>0, deci x1=(3-5)/(2·2)=-2/4=-1/2; x2=(3+5)/4=2.

Raspuns: S={-1/2; 2}

Explicație pas cu pas: