Question

3 задание решите плез

Answer 1

Ответ:

$\frac{3}{y^{2}}-\frac{1}{6};$ $-\frac{4}{a+5};$ $\frac{10y}{y^{2}-100};$ $\frac{18c}{2c-3};$

Объяснение:

$3. 1) \frac{(3-2y)*6-(y-12)*y}{6y^{2}}=\frac{18-12y-y^{2}+12y}{6y^{2}}=\frac{18-y^{2}}{6y^{2}}=\frac{18}{6y^{2}}-\frac{y^{2}}{6y^{2}}=\frac{3}{y^{2}}-\frac{1}{6};$

$2) \frac{20}{a*(a+5)}-\frac{4*(a+5)}{a*(a+5)}=\frac{20-4a-20}{a*(a+5)}=\frac{-4a}{a*(a+5)}=-\frac{4}{a+5};$

$3) \frac{y}{y-10}-\frac{y^{2}}{(y-10)(y+10)}=\frac{y(y+10)}{(y-10)(y+10)}-\frac{y^{2}}{(y-10)(y+10)}=\frac{y^{2}+10y-y^{2}}{(y-10)(y+10)}=\frac{10y}{y^{2}-100};$

$4) \frac{12c^{2}-6c*(2c-3)}{2c-3}=\frac{12c^{2}-12c^{2}+18c}{2c-3}=\frac{18c}{2c-3};$

Answer 2

Ответ:

$\frac{y}{y-10\\} - \frac{y^{2} }{y^{2}-100 }= \\\\\=\frac{y}{y-10} - \frac{y^{2} }{(y-10)(y+10)}=\\\\\=\frac{y(y+10)-y^{2} }{(y-10)(y+10)} = \frac{y^{2}+10y-y^{2} }{(y-10)(y+10)} = \frac{10y}{y^{2}-100 }$

Ответ: $\frac{10y}{y^{2} -100}$