Đáp án:

 a)đk: \(x\neq0; x\neq10; x\neq-10\)

b)\(\frac{10}{x}\)

c)\(\frac{1}{2004}\)

Giải thích các bước giải:

a)đk: \(x\neq0; x\neq10; x\neq-10\)

b)\((\frac{5x+2}{x^{2}-10x}+\frac{5x-2}{x^{2}+10x})\cdot \frac{x^{2}-100}{x^{2}+4}\)
=\((\frac{5x+2}{x(x-10)}+\frac{5x-2}{x(x+10)})\cdot \frac{(x+10)(x-10)}{x^{2}+4}\)
=\((\frac{(5x+2)(x+10)+(5x-2)(x-10)}{x(x-10)(x+10)})\cdot \frac{(x+10)(x-10)}{x^{2}+4}\)
=\((\frac{5x^{2}+52x+20+5x^{2}-52x+20}{x(x-10)(x+10)})\cdot \frac{(x+10)(x-10)}{x^{2}+4}\)
=\(\frac{10x^{2}+40}{x(x-10)(x+10)}\cdot \frac{(x+10)(x-10)}{x^{2}+4}\)
=\(\frac{10(x^{2}+4)}{x(x-10)(x+10)}\cdot \frac{(x+10)(x-10)}{x^{2}+4}\)
=\(\frac{10}{x}\)

c)Tại x=20040⇒\( \frac{10}{20040}=\frac{1}{2004}\)