a, \(ĐKXĐ:\left\{{}\begin{matrix}x^2-25\ne0\\5-x\ne0\\x+5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
\(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{\left(x-5\right)}\)
b, Để A đạt giá trị nguyên
\(\Leftrightarrow\dfrac{-4}{x-5}\) đạt giá trị nguyên
\(\Leftrightarrow4⋮x-5\)
\(\Leftrightarrow x-5\inƯ\left(4\right)=\left\{1;-1;4;-4;2;-2\right\}\)
| \(x-5\) | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 6 | 4 | 7 | 3 | 9 | 1 |
Vì tất cả các giá trị trên đều thỏa mãn ĐKXĐ
Vậy \(x\in\left\{1;3;4;6;7;9\right\}\) thì A đạt giá trị nguyên
\(\text{a) }ĐKXĐ:x\ne\pm5\)
Với \(x\ne\pm5\), ta có:
\(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\\=\dfrac{2x}{\left(x+5\right)\left(x-5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\\ =\dfrac{2x}{\left(x+5\right)\left(x-5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{2x-5x-25-x+5}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{-4x-20}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{-4\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{4}{5-x}\)
Vậy \(A=\dfrac{4}{5-x}\) với \(x\ne\pm5\)
b) Với \(x\ne\pm5\)
Để A nhận giá trị nguyên
thì \(\Rightarrow\dfrac{4}{5-x}\in Z\)
\(\Rightarrow4\text{ }⋮\text{ }5-x\\ \Rightarrow5-x\inƯ_{\left(4\right)}\)
Mà \(Ư_{\left(4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)
Lập bảng giá trí:
| \(5-x\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) |
| \(x\) | \(9\left(TM\right)\) | \(7\left(TM\right)\) | \(6\left(TM\right)\) | \(4\left(TM\right)\) | \(3\left(TM\right)\) | \(1\left(TM\right)\) |
Vậy với \(x=\left\{9;7;6;4;3;1\right\}\)
thì A nhận giá trị nguyên.
a)
\(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\\ =\dfrac{2x}{x^2-25}-\dfrac{5\left(x+2\right)}{x^2-25}-\dfrac{x-5}{x^2-25}\\ =\dfrac{2x-5x-10-x+5}{x^2-25}\\ =\dfrac{-4x-5}{x^2-25}\)
