Question

помогите срочно очень нужно в зарание спасибо !​

Answer 1

$\displaystyle1)\frac{7x}{21y}=\frac{x}{3y}\\\\\\2) \frac{16ab}{4ac}=\frac{4b}{c}\\\\\\3)\frac{25n^3}{15n^6}=\frac{5}{3n^3} \\\\\\4)\frac{8ab^3c^2}{16a^2bc^3}=\frac{b^2}{2ac}\\\\\\5)\frac{48a^5b^7}{32a^3b^8}=\frac{3a^2}{2b}\\\\\\6)\frac{34m^9n^3}{54m^4n^7}= \frac{17m^5}{27n^4}$

$\displaystyle1)\frac{9c+27}{9c}=\frac{9(c+3)}{9c}=\frac{c+3}{c}\\\\\\2)\frac{6a-18b}{7a-21b}=\frac{6(a-3b)}{7(a-3b)}=\frac{6}{7}\\\\\\3)\frac{a^2-64}{3a+24}=\frac{a^2-8^2}{3(a+8)}=\frac{(a-8)(a+8)}{3(a+8)}=\frac{a-8}{3}\\\\\\4)\frac{20x^2-5x}{6-24x}=\frac{20x^2-5x}{-24x+6} =\frac{5x(4x-1)}{-6(4x-1)}=\frac{5x}{-6}=-\frac{5x}{6} \\\\\\5)\frac{x^2-25}{x^2-10x+25}=\frac{x^2-5^2}{x^2-2*5x+5^2}=\frac{(x-5)(x+5)}{(x-5)^2}=\frac{(x+5)}{(x-5)}$

$\displaystyle6) \frac{a^5-a^7}{a^6-a^4}=\frac{a^5-a^7}{-a^4+a^6}=\frac{a^5(1-a^2)}{-a^4(1-a^2)}=\frac{a^5}{-a^4}=\frac{a}{-1}=-a\\\\\\7)\frac{a^3-216}{2a-12} =\frac{a^3-6^3}{2(a-6)}=\frac{(a-6)(a^2+6a+6^2)}{2(a-6)}=\frac{a^2+6a+6^2}{2} =\frac{(a+6)^2}{2}\\\\\\8)\frac{3m^2-6m+12}{6m^3+48}=\frac{3(m^2-2m+4)}{6(m^3+8)}=\frac{3(m^2-2m+4)}{6(m^3+2^3)}=\\\\\\\frac{3(m^2-2m+4)}{6(m+2)(m^2-2m+2^2)}=\frac{3(m^2-2m+4)}{6(m+2)(m^2-2m+4)}=\frac{3}{6(m+2)}$

$\displaystyle9)\frac{am-mb-6a+6b}{m^2-36}=\frac{m(a-b)-6(a-b)}{m^2-6^2}=\frac{(m-6)(a-b)}{(m-6)(m+6)}=\frac{a-b}{m+6}$

С помощью формул: