Question

помогите срочно очень мало времни) заранее спасибо

Answer 1

№1

$1) \frac{2-12m}{3m} \\ \\ 2) \frac{3-12c}{2c} \\ \\ 3) \frac{5 a^{3} + 3}{ a^{2} } \\ \\ 4) \frac{7b^{4} +2}{ b^{3} }$

№2

$1) \frac{z-8z}{4(x+y)} = \frac{-7z}{4(x-y)} \\ \\ 2) \frac{2c-9c}{3(a-b)} = \frac{-7c}{3(a-b)} \\ \\ 3) \frac{6b+28b}{105(b+c)}= \frac{34b}{105(b+c)} \\ \\ 4) \frac{4n+9n}{72(m+n)}= \frac{13n}{72(m+n)}$

№3

$1) \frac{3}{x(x-2)} + \frac{1}{2y(x-2)} = \frac{6y+x}{2xy(x-2)} \\ \\ 2) \frac{2}{c(3c-1)} + \frac{1}{2b(3c-1)} = \frac{4b+c}{2bc(3c-1)}$

№4

$1) \frac{a}{b(a-b)} - \frac{2a}{a-b} = \frac{a-2ab}{b(a-b)} \\ \\ 2) \frac{4y-x}{y(x-y)} \\ \\ 3) \frac{5b^{2} + 3a^{2} }{ab(x-y)} \\ \\ 4) \frac{2b^{2} +4c^{2} }{cb(k-n)}$

№5

$1) \frac{2x}{(3+x)(3-x)} - \frac{1}{2(x-3)} = \frac{5x+3}{2(3+x)(3-x)} \\ \\ 2) \frac{1}{3(a-2)} - \frac{3a}{4-a^{2}} = \frac{7a+2}{3(a-2)(a+2)} \\ \\ 3) \frac{2}{(2c+5)(2c-5)} + \frac{3c}{3(5-2c)} = \frac{6-6c^{2} -15c }{3(2c+5)(2c-5)} = \frac{2-2c^{2} -3c}{(2c+5)(2c-5)} \\ \\ 4) \frac{2x}{2(4-9x)} -\frac{3}{(9x-4)(9x+4)} = \frac{18x^{2} +8x+6}{2(4-9x)(4+9x)} = \frac{9 x^{2}+4x-3}{(4-9x)(4+9x)}$