Question

СОРЧНО! решите эти два уравнения .

Answer 1

Объяснение:

$\large \boldsymbol{} \sf ODZ: x\neq \pm 3$

$\large \boldsymbol {} \displaystyle b) \ \ \frac{2y-2}{y+3} +\frac{y+3}{y-3} =5 \ \ | \times (y-3) (y+3) \\\\\\ (2y-2)(y-3) +(y+3) (y+3) =5\underbrace{(y-3) (y+3) }_{(y^2-9 )} \\\\\\ 2y^2-2y-\boldsymbol {6y\!\!\!\!/}+6+y^2+\boldsymbol {6y\!\!\!\!/}+9 = 5y^2-45 \\\\\\ 3y^2-2y+9 = 5y^2-45 \\\\\\ 5y^2-3y^2+2y-45-9-6 =0 \\\\\\ 2y^2+2y-60=0 \ \ | \div 2 \\\\ y^2+y-30 =0 \\\\ \left \{ {{y_1+y_2=-1} \atop {y_1\cdot y_2=-30 }} \right. = y _1=-6 \ ; \ y_2=5 \\\\\\ Otvet : \boxed{y _1=-6 \ ; \ y_2=5 }$

$c) \ \ \large \boldsymbol {} \displaystyle {} \frac{x^2+4x}{x+2} =\frac{2x}{3} \ \ \sf {ODZ : x\neq -2 }$

$\large \boldsymbol {} \displaystyle 3( x^2+4x) =2x(x+2) \\\\ 3x^2 +12x =2x^ 2+4x \\\\ 3x^ 2-2x^2 +12x-4x=0 \\\\x^2 +8x=0 \\\\ x(x+8 ) =0 = x_1 =0 \ \ ; \ \ x_2=-8 \\\\ Otvet : \boxed{x_ 1=0 \ \ ; \ \ x_2=-8 }$