\(A=\frac{x^2}{x^2-4x}+\frac{6}{6-3x}+\frac{1}{x^2}\)
a) Điều kiện xác định: \(\left\{{}\begin{matrix}x^2-4x\ne0\\6-3x\ne0\\x^2\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x.\left(x-4\right)\ne0\\3.\left(2-x\right)\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ne0\\x-4\ne0\end{matrix}\right.\\2-x\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne4\\x\ne2\end{matrix}\right.\)
Vậy để A được xác định thì \(x\ne0;x\ne4\) và \(x\ne2.\)
Rút gọn:
\(A=\frac{x^2}{x^2-4x}+\frac{6}{6-3x}+\frac{1}{x^2}\left(x\ne0;x\ne4;x\ne2\right)\)
\(A=\frac{x^2}{x.\left(x-4\right)}+\frac{6}{3.\left(2-x\right)}+\frac{1}{x^2}\)
\(A=\frac{3x.x^2.\left(2-x\right)}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{6x^2.\left(x-4\right)}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{3.\left(x-4\right).\left(2-x\right)}{3x^2.\left(x-4\right).\left(2-x\right)}\)
\(A=\frac{3x^3.\left(2-x\right)}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{6x^3-24x^2}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{\left(3x-12\right).\left(2-x\right)}{3x^2.\left(x-4\right).\left(2-x\right)}\)
\(A=\frac{6x^3-3x^4}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{6x^3-24x^2}{3x^2.\left(x-4\right).\left(2-x\right)}+\frac{6x-3x^2-24+12x}{3x^2.\left(x-4\right).\left(2-x\right)}\)
\(A=\frac{6x^3-3x^4+6x^3-24x^2+18x-3x^2-24}{3x^2.\left(x-4\right).\left(2-x\right)}\)
\(A=\frac{12x^3-3x^4-27x^2+18x-24}{3x^2.\left(x-4\right).\left(2-x\right)}\)
\(A=\frac{3.\left(4x^3-x^4-9x^2+6x-8\right)}{3x^2.\left(x-4\right).\left(2-x\right)}\)
Đoạn này chịu rồi.
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