Question

Можно с решением пожалуйста

Answer 1

$1)\frac{5a-c}{8p} *\frac{7}{5a-c}=\frac{7}{8p}\\\\2)\frac{a^{2}-2ab }{x^{2}+xy } *\frac{x+y}{a-2b} =\frac{a(a-2b)*(x+y)}{x(x+y)*(a-2b)}=\frac{a}{x}\\\\3)\frac{7a^{15}}{c^{7}}:\frac{21a^{16}}{c^{8}}=\frac{7a^{15}}{c^{7}}*\frac{c^{8} }{21a^{16} }=\frac{c}{3a} \\\\4)\frac{x^{2}-c^{2}}{a-c}:\frac{x+c}{a^{2}-c^{2}}=\frac{(x-c)(x+c)}{a-c}*\frac{(a-c)(a+c)}{x+c}=(x-c)(a+c)$

$5)\frac{a^{2}-b^{2}}{a^{2}-2ax+x^{2}} :\frac{a-b}{a-x}=\frac{(a-b)(a+b)}{(a-x)^{2}}*\frac{a-x}{a-b}=\frac{a+b}{a-x}\\\\6)\frac{8y}{y^{2}-a^{2}} *(y+a)=\frac{8y*(y+a)}{(y+a)(y-a)}=\frac{8y}{y-a}\\\\7)\frac{m^{2}-4n^{2}}{m+n} :(m-2n)=\frac{(m-2n)(m+2n)}{m+n}*\frac{1}{m-2n}=\frac{m+2n}{m+n}\\\\8)\frac{p^{2}-4c^{2}}{a^{2}+10ab+25b^{2}}*\frac{a^{2}-25b^{2}}{2c-p}=\frac{(p-2c)(p+2c)}{(a+5b)^{2}}*\frac{(a-5b)(a+5b)}{2c-p} =-\frac{(p+2c)(a-5b)}{a+5b}=\frac{(p+2c)(5b-a)}{a+5b}$