Question

Calculati:a) $\frac{2-3x}{x^{2}-x }+ \frac{2}{x-1}+ \frac{5}{3x}$
b)$\frac{x-1}{x-2}+ \frac{x+3}{x+2}+ \frac{2}{x^{2}-4 };$

Answer 1

[tex]a)\frac{2-3x}{x^2-x}+\frac{2}{x-1}+\frac{5}{3x}=\\ \frac{2-3x}{x(x-1)}+\frac{2}{x-1}+\frac{5}{3x}=\\ \frac{3(2-3x)+6x+5(x-1)}{3x(x-1)}=\\ \frac{6-9x+6x+5x-5}{3x(x-1)}=\frac{1+2x}{3x(x-1)},x\notin\{0,1\} \\ b)\frac{x-1}{x-2}+\frac{x+3}{x+2}+\frac{2}{x^2-4}=\\ \frac{x-1}{x-2}+\frac{x+3}{x+2}+\frac{2}{(x-2)(x+2)}=\\ \frac{(x-1)(x+2)+(x+3)(x-2)+2}{(x-2)(x+2)}=\\ \frac{x^2-x+2x-2+x^2-2x+3x-6+2}{(x-2)(x+2)}=\frac{2x^2+2x-6}{(x-2)(x+2)},x\notin \{2,-2\}[/tex]