x/x-1+2x+6/x^2-3x-2/x^2-x): x^2-x+4/x^2-4x+3 tìm đk mình cần gấp trc 7h 05/12/2021 Bởi Skylar x/x-1+2x+6/x^2-3x-2/x^2-x): x^2-x+4/x^2-4x+3 tìm đk mình cần gấp trc 7h
Đáp án: \(x \ne \left\{ {0;1;3} \right\}\) Giải thích các bước giải: \(\begin{array}{l}\left( {\dfrac{x}{{x – 1}} + \dfrac{{2x + 6}}{{{x^2} – 3x}} – \dfrac{2}{{{x^2} – x}}} \right):\dfrac{{{x^2} – x + 4}}{{{x^2} – 4x + 3}}\\DK:\left\{ \begin{array}{l}{x^2} – 3x \ne 0\\{x^2} – x \ne 0\\{x^2} – 4x + 3 \ne 0\end{array} \right.\\ \to \left\{ \begin{array}{l}x\left( {x – 3} \right) \ne 0\\x\left( {x – 1} \right) \ne 0\\\left( {x – 1} \right)\left( {x – 3} \right) \ne 0\end{array} \right. \to \left\{ \begin{array}{l}x \ne 0\\x \ne 1\\x \ne 3\end{array} \right.\\ \to x \ne \left\{ {0;1;3} \right\}\end{array}\) Bình luận
Đáp án:
\(x \ne \left\{ {0;1;3} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {\dfrac{x}{{x – 1}} + \dfrac{{2x + 6}}{{{x^2} – 3x}} – \dfrac{2}{{{x^2} – x}}} \right):\dfrac{{{x^2} – x + 4}}{{{x^2} – 4x + 3}}\\
DK:\left\{ \begin{array}{l}
{x^2} – 3x \ne 0\\
{x^2} – x \ne 0\\
{x^2} – 4x + 3 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x\left( {x – 3} \right) \ne 0\\
x\left( {x – 1} \right) \ne 0\\
\left( {x – 1} \right)\left( {x – 3} \right) \ne 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne 0\\
x \ne 1\\
x \ne 3
\end{array} \right.\\
\to x \ne \left\{ {0;1;3} \right\}
\end{array}\)