Lời giải:
a. ĐKXĐ: \(\left\{\begin{matrix}
x^2-10x\neq 0\\
x^2+10x\neq 0\\
x^2+4\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x(x-10)\neq 0\\
x(x+10)\neq 0\\
x^2+4\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0; x\neq \pm 10\)
b.
\(P=\left[\frac{(5x+2)(x+10)}{x(x-10)(x+10)}+\frac{(5x-2)(x-10)}{x(x+10)(x-10)}\right].\frac{(x-10)(x+10)}{x^2+4}\)
\(=\frac{(5x+2)(x+10)+(5x-2)(x-10)}{x(x-10)(x+10)}.\frac{(x-10)(x+10)}{x^2+4}\)
\(=\frac{10x^2+40}{x}.\frac{1}{x^2+4}=\frac{10(x^2+4)}{x}.\frac{1}{x^2+4}=\frac{10}{x}\)
c.
Khi $x=20040$ nên $P=\frac{10}{20040}=\frac{1}{2004}$
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