Решите плиз,тема: числовое значение рационального выражения. Даю 93балла

Question 1

Question 2

Не видно значение переменной у в примере №3.

Question 3

Решите задачу:

$1)\; \; m=0,5=\frac{1}{2}\; ,\; \; n= \frac{2}{3} \\\\\frac{3m^2+6mn+3n^2}{6n^2-6m^2}=\frac{3(m^2+2mn+n^2)}{6(n^2-m^2)}=\frac{3(m+n)^2}{6(n-m)(n+m)}= \frac{m+n}{2(n-m)}=\\\\=\frac{ \frac{2}{2}+\frac{2}{3} }{2( \frac{2}{3}-\frac{1}{2} )} = \frac{\frac{10}{6}}{2\cdot \frac{1}{6}} = \frac{\frac{5}{3}}{\frac{1}{3}} =\frac{5\cdot 3}{3}=5$

$2)\; \; b=0,25=\frac{1}{4}\; ,\; \; c=\frac{1}{3}\\\\\frac{2c^2-2b^2}{4b^2-8bc+4c^2}= \frac{2(c-b)(c+b)}{4(b-c)^2}=\frac{c+b}{2(c-b)} = \frac{\frac{1}{4}+\frac{1}{3}}{2(\frac{1}{3}-\frac{1}{4})}=\\\\=\frac{7}{12\cdot 2\cdot \frac{1}{12}}=\frac{7}{2}=3,5$

$4)\; \; x=5,125\\\\\frac{x^2+25}{(x-5)^3}+\frac{10x}{(5-x)^3}= \frac{x^2+25}{(x-5)^3}-\frac{10x}{(x-5)^3} =\frac{x^2-10x+25}{(x-5)^3}=\\\\=\frac{(x-5)^2}{(x-5)^3} =\frac{1}{x-5}=\frac{1}{5,125-5}=\frac{1}{0,125}=\frac{1}{\frac{125}{1000}}=\frac{1000}{125}=8$

$3)\; \; x=0,35\; ,\; y=???\\\\\frac{4xy}{y^2-x^2}:\Big (\frac{1}{y^2-x^2}+\frac{1}{x^2+2xy+y^2}\Big )=\\\\=\frac{4xy}{(y-x)(y+x)}:\Big(\frac{1}{(y-x)(y+x)}+ \frac{1}{(x+y)^2}\Big )=\\\\=\frac{4xy}{(y-x)(x+y)}:\frac{x+y+y-x}{(y-x)(x+y)^2}= \frac{4xy}{(y-x)(x+y)}\cdot \frac{(y-x)(x+y)^2}{2y}=\\\\= \frac{2x(x+y)}{1}=2x(x+y)$