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[tex]\displaystyle 2)\ a)\ \frac{7}{a}+\frac{a-7}{a+5}=\frac{7(a+5)+(a-7)\cdot a}{a\, (a+5)}=\frac{7a+35+a^2-7a}{a\, (a+5)}=\frac{a^2+35}{a\, (a+5)}\\\\\\b)\ \ \frac{2x^2}{x^2-4}-\frac{2x}{x+2}=\frac{2x^2}{(x-2)(x+2)}-\frac{2x}{x+2}=\frac{2x^2-2x(x-2)}{(x-2)(x+2)}=\\\\\\=\frac{2x^2-2x^2+4x}{(x-2)(x+2)}=\frac{4x}{x^2-4}[/tex]

[tex]4)\ \ \dfrac{x}{y}=f\ \ ,\ \ \dfrac{y}{z}=\dfrac{1}{f}\ \ \ \ \Rightarrow \ \ \ \ \ \ \dfrac{x}{z}=\dfrac{x}{y}\cdot \dfrac{y}{z}=f\cdot \dfrac{1}{f}=1[/tex]