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Решите задачу:

$=\frac{5}{x-7}-\frac{2}{x}-\frac{3x+28}{x^{2}-7^2} =\frac{5}{x-7}-\frac{2}{x}-\frac{3x+28}{(x-7)(x+7)}=$

$=\frac{5*x*(x+7)}{(x-7)*x*(x+7)}-\frac{2*(x-7)*(x+7)}{x*(x-7)*(x+7)}-\frac{x*(3x+28)}{x*(x-7)(x+7)}=$

$=\frac{5x(x+7)}{x(x-7)(x+7)}-\frac{2(x^2-49)}{x*(x-7)*(x+7)}-\frac{3x^2+28x}{x(x-7)(x+7)}=$

$=\frac{5x^2+35x}{x(x-7)(x+7)}-\frac{2x^2-98}{x*(x-7)*(x+7)}-\frac{3x^2+28x}{x(x-7)(x+7)}=$

$=\frac{5x^2+35x-(2x^2-98)-(3x^2+28x)}{x(x-7)(x+7)} =\frac{5x^2+35x-2x^2+98-3x^2-28x}{x(x-7)(x+7)}=$

$=\frac{5x^2-2x^2-3x^2+35x-28x+98}{x(x-7)(x+7)} =\frac{7x+98}{x(x-7)(x+7)}= \frac{7x^2+98}{x(x^2-49)}= \frac{7x^2+98}{x^3-49x}$