Question

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Answer 1

$1)\; \; \dfrac{x-3,5}{10-5x}\; \; ,\; \; \; 10-5x\ne 0\; \; \; \Rightarrow \; \; \; \boxed {\; x\ne 2\; }\\\\\\2)\; \; \dfrac{8x^5y^2}{16y^4x^3}=\dfrac{x^2}{2y^2}\; \; \; ,\; \; \; \; \; \dfrac{x^2-16}{4x+16}=\dfrac{(x-4)(x+4)}{4\, (x+4)}=\dfrac{x-4}{4}$

$3)\; \; \dfrac{x+3}{2}-\dfrac{x-5}{2}=\dfrac{x+3-x+5}{2}=4\\\\\\\dfrac{3}{x}+\dfrac{x-4}{x+3}=\dfrac{3x+9+x^2-4x}{x(x+3)}=\dfrac{x^2-x+9}{x(x+3)}\\\\\\\dfrac{3a-3b}{3a^2}\cdot \dfrac{a^2}{a^2-b^2}=\dfrac{3(a-b)}{3a^2}\cdot \dfrac{a^2}{(a-b)(a+b)}=\dfrac{1}{a+b}$

$\dfrac{m^2-9}{n}:\dfrac{m^2-6m+9}{5n^2}=\dfrac{(m-3)(m+3)}{n}\cdot \dfrac{5n^2}{(m-3)^2}=\dfrac{5n\, (m+3)}{m-3}$