Question

СОЧ НУЖНО СЕЙЧАС СДАТЬ Выполните сложение и вычитание алгебраических дробей.

Answer 1

1. $\frac{8y-1}{18y}$

2. $\frac{y^{5} +8y}{9-y^{2}}$

Объяснение:

1. $\frac{7y+4}{8y}-\frac{2y+3}{6y} = \frac{7y+4}{8y}*2-\frac{2y+3}{6y}*3=$$\frac{(14y+8)-(6y+9)}{18y} = \frac{14y+8-6y-9}{18y}=\frac{8y-1}{18y}$

2. $\frac{2-3y}{y^{2}-9} - \frac{5-2y}{3-y} = \frac{2-3y}{(y-3)*(y+3)} - \frac{5-2y}{3-y} =$$-\frac{2+3y}{(3-y)*(3+y)} -\frac{5-2y}{3-y} * (y-3) = -\frac{2+3y}{(3-y)*(3+y)} - \frac{(5-2y)* (3-y)}{(3-y)*(3+y)} =$  $\frac{(2+3y)-(5y+15-2y^{2}-6y) }{9-y^{2} } = \frac{(2+3y)- (2y^{2}-y+15)}{9-y^{2} } = \frac{4y^{2}-2y+30+6y^{3}-3y^{2}+45y}{9-y^{2}} = \frac{6y^{3} +y^{2}+43y+30 }{9-y^{2}}=$$\frac{(6y^{3}+30) +(y^{2} +43y)}{9-y^{2}} =\frac{6(y^{3}+5)+y(y+43)}{9-y^{2}} = \frac{6y (y^{3}+5+y+43) }{9-y^{2} } = \frac{6y(y^{4}+48) }{9-y^{2} } = \frac{y(y^{4}+8) }{9-y^{2} }= \frac{y^{5} +8y}{9-y^{2}}$