Question

ПОЖАЛУЙСТА ОЧЕНЬ СРОЧНО! РЕШИТЕ 3 УРАВНЕНИЙ.ВСЕ УРАВНЕНИЯ.​

Answer 1

$1)\ \ \dfrac{3x+1}{x+2}-\dfrac{x-1}{x-2}=1\ \ ,\ \ \ ODZ:\ x\ne \pm 2\ ,\\\\\dfrac{(3x+1)(x-2)-(x+2)(x-1)-(x+2)(x-2)}{(x+2)(x-2)}=0\\\\\dfrac{(3x^2-5x-2)-(x^2+x-2)-(x^2-4)}{(x+2)(x-2)}=0\\\\\dfrac{x^2-6x+4}{(x+2)(x-2)}=0\ \ \ \to \ \ \ x^2-6x+4=0\ \ ,\ \ x=3\pm \sqrt{5}\\\\x_1=3-\sqrt5\ \ ,\ \ x_2=3+\sqrt5$

$2)\ \ \dfrac{2y-2}{y+3}+\dfrac{y+3}{y-3}=5\ \ ,\ \ \ ODZ:\ \ y\ne \pm 3\ ,\\\\\dfrac{(2y-2)(y-3)+(y+3)^2-5(y+3)(y-3)}{(y+3)(y-3)}=0\\\\\dfrac{(2y^2-8y+6)+(y^2+6y+9)-(5y^2-45)}{(y+3)(y-3)}=0\\\\\dfrac{-2y^2-2y+60}{(y+3)(y-3)}=0\ \ \ \to \ \ \ y^2+y-30=0\ \ ,\ \ D=121\ ,\\\\\\y_1=-6\ ,\ y_2=5$

$3)\ \ \ \dfrac{4}{9y^2-1}-\dfrac{4}{3y+1}=\dfrac{5}{1-3y}\ \ ,\ \ \ ODZ:\ y\ne \pm \dfrac{1}{3}\ ,\\\\\dfrac{4-4(3y-1)+5(3y+1)}{(3y-1)(3y+1)}=0\\\\\dfrac{4-12y+4+15y+5}{(3y-1)(3y+1)}=0\ \ ,\ \ \dfrac{3y+13}{(3y-1)(3y+1)}=0\ \ \to \\\\\\3y+13=0\ \ ,\ \ \ y=-\dfrac{13}{3}\ \ ,\ \ y=-4\dfrac{1}{3}$