Question

Помогите очень срочно пожалуйста(((Сократите дробь:а)б)в)г)д)е)ж)з)и)

Answer 1

$\frac{x}{xy-x}=\frac{x}{x(y-1)}=\frac{1}{y-1}$
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$\frac{a^2b+ab}{ab}=\frac{ab(a+1)}{ab}=a+1$
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$\frac{2mnp}{2m^2p-6mp}=\frac{2mp*n}{2mp*m-2mp*3}=\\\\\frac{2mp*n}{2mp(m-3)}=\frac{n}{m-3}$
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$\frac{5a^2}{a^4+a^2}=\frac{5*a^2}{a^2*a^2+a^2*1}=\\\\\frac{5*a^2}{a^2(a^2+1)}=\frac{5}{a^2+1}$
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$\frac{b^5-b^4}{b^5}=\frac{b^4*b-b^4*1}{b^4*b}=\\\\\frac{b^4(b-1)}{b^4*b}=\frac{b-1}{b}$
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$\frac{n^2-n-1}{n^4-n^3-n^2}=\frac{(n^2-n-1)*1}{n^2*n^2-n^2*n-n^2*1}=\\\\\frac{(n^2-n-1)*1}{n^2(n^2-n-1)}=\frac{1}{n^2}$
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$\frac{3ac+6a}{ac^2+2ac}=\frac{3a*c+3a*2}{ac*c+ac*2}=\\\\\frac{3a(c+2)}{ac(c+2)}=\frac{3}{c}$
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$\frac{4a-2a^2}{4b-2ab}=\frac{2a*2-2a*a}{2b*2-2b*a}=\\\\\frac{2a(2-a)}{2b(2-a)}=\frac{a}{b}$
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$\frac{ac^2-dc^2}{c(a^2-b^2)}=\frac{c*(ac-dc)}{c(a^2-b^2}=\\\\\frac{ac-dc}{a^2-b^2}$
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$\frac{ac^2-bc^2}{c(a^2-b^2)}=\frac{c^2(a-b)}{c(a-b)(a+b)}=\\\\\frac{c}{a+b}$