39.15 помогите срочно

Решите задачу:

$1) 2 - \frac{a}{3} - \frac{b}{4} = \frac{2*12 - 4a - 3b}{12} = \frac{24-4a-3b}{12} \\ \\ 2)12 - \frac{1}{a} - \frac{1}{b} = \frac{12ab -1*b -1*a}{ab} = \frac{12ab-b-a}{ab} \\ \\ 3) \frac{x-2}{2} -4- \frac{x-3}{3} = \frac{3(x-2)-4*6-2(x-3)}{6} = \frac{3x-6-24-2x+6}{6} = \frac{x-24}{6} \\ \\ 4)4a - \frac{a-1}{4} - \frac{a+2}{3} = \frac{4a*12-3(a-1)-4(a+2)}{12} = \frac{48a-3a+3-4a-8}{12} = \frac{41a-5}{12}$

$5) \frac{a+b}{4} -a+b = \frac{a+b -4a+4b}{4} = \frac{5b-3a}{4} \\ \\ 6)2a+3b- \frac{a^2+b^2}{a} = \frac{2a^2+3ab -(a^2+b^2)}{a} = \frac{2a^2+3ab -a^2-b^2}{a} = \frac{a^2+3ab-b^2}{a} \\ \\ 7)3x+3b- \frac{x^2+2y^2}{x} = \frac{3x^2+3bx - (x^2+2y^2)}{x} = \frac{3x^2+3bx-x^2-2y^2}{x} = \frac{2x^2+3bx-2y^2}{x} \\ \\ 8)5a - 3b- \frac{3a^2-b^2}{a} = \frac{5a^2-3ab -(3a^2-b^2)}{a} = \frac{5a^2-3ab -3a^2+b^2}{a} = \frac{2a^2-3ab +b^2}{a} \\ \\$