Question

упростить надо
помогите пожалуйста надо на 5​

Answer 1

$\tt\displaystyle 1)\ \frac{a}{6} +\frac{b}{6} =\boxed{\frac{a+b}{6}}$

$\tt\displaystyle 2)\ \frac{x+4y}{12} +\frac{2x+5y}{12} = \frac{x+4y+2x+5y}{12} =\frac{3x+9y}{12} =\frac{3(x+3y)}{12}=\boxed{\frac{x+3y}{4}}$

$\tt\displaystyle 3)\ \frac{a-8}{a^2-25}+ \frac{13}{a^2-25} =\frac{a-8+13}{a^2-25} =\frac{a+5}{(a-5)(a+5)} =\boxed{\frac{1}{a-5}}$

$\tt\displaystyle 4)\ \frac{3c}{c^2-5c} -\frac{10+c}{c^2-5c} =\frac{3c-10-c}{c^2-5c} =\frac{2c-10}{c^2-5c} =\frac{2(c-5)}{c(c-5)} =\boxed{\frac{2}{c}}$

$\tt\displaystyle 5)\ \frac{x+2}{x-2} -\frac{x}{2-x} =\frac{x+2}{x-2}+\frac{x}{x-2} =\frac{x+2+x}{x-2} =\boxed{\frac{2x+2}{x-2}}$

$\tt\displaystyle 6)\ \frac{7-4y}{(y-2)^2} -\frac{8-5y}{(2-y)^2} =\frac{(7-4y)(2-y)^2-(8-5y)(y-2)^2}{(y-2)^2(2-y)^2}=\\\\\\=\frac{(7-4y)(4-4y+y^2)-(8-5y)(y^2-4y+4)}{(y-2)^2(2-y)^2}=\\\\\\=\frac{28-16y-28y+16y^2+7y^2-4y^3-(8y^2-5y^3-32y+20y^2+32-20y)}{(y-2)^2(2-y)^2}=\\\\\\=\frac{y^3-5y^2+8y-4}{(y-2)^2(2-y)^2}=\frac{y^3-(y^2+4y^2)+(4y+4y)-4}{(y-2)^2(2-y)^2}=\\\\\\=\frac{y^3-y^2-4y^2+4y+4y-4}{(y-2)^2(2-y)^2}=\frac{(y^3-y^2)+(-4y^2+4y)+(4y-4)}{(y-2)^2(2-y)^2}=$

$\tt\displaystyle =\frac{y^2(y-1)+4y(-y+1)+4(y-1)}{(y-2)^2(2-y)^2}=\frac{y^2(y-1)-4y(y-1)+4(y-1)}{(y-2)^2(2-y)^2}=\\\\\\=\frac{(y-1)(y^2-4y+4)}{(y-2)^2(2-y)^2}=\frac{(y-1)(y-2)^2}{(2-y)^2(y-2)^2}=\boxed{\frac{y-1}{(2-y)^2}}$

или

$\tt\displaystyle 6)\ \frac{7-4y}{(y-2)^2} -\frac{8-5y}{(2-y)^2} = \frac{7-4y}{(y-2)^2} -\frac{8-5y}{(-(-2+y))^2} = \frac{7-4y}{(y-2)^2} -\frac{8-5y}{(-(y-2))^2} =\\\\\\= \frac{7-4y}{(y-2)^2} -\frac{8-5y}{(y-2)^2} =\frac{7-4y-8+5y}{(y-2)^2} =\boxed{\frac{y-1}{(y-2)^2} }$