Căn của cả x^2 – 9 / x – 1 có nghĩa khi 03/08/2021 Bởi Elliana Căn của cả x^2 – 9 / x – 1 có nghĩa khi
Đáp án: \(\left[ \begin{array}{l}x \ge 3\\ – 3 \le x < 1\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\sqrt {\dfrac{{{x^2} – 9}}{{x – 1}}} \\DK:\left\{ \begin{array}{l}\dfrac{{{x^2} – 9}}{{x – 1}} \ge 0\\x – 1 \ne 0\end{array} \right.\\ \to \left\{ \begin{array}{l}x \ne 1\\\left[ \begin{array}{l}\left\{ \begin{array}{l}{x^2} – 9 \ge 0\\x – 1 > 0\end{array} \right.\\\left\{ \begin{array}{l}{x^2} – 9 \le 0\\x – 1 < 0\end{array} \right.\end{array} \right.\end{array} \right.\\ \to \left\{ \begin{array}{l}x \ne 1\\\left[ \begin{array}{l}\left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 3\\x \le – 3\end{array} \right.\\x > 1\end{array} \right.\\\left\{ \begin{array}{l} – 3 \le x \le 3\\x < 1\end{array} \right.\end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}x \ge 3\\ – 3 \le x < 1\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x \ge 3\\
– 3 \le x < 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {\dfrac{{{x^2} – 9}}{{x – 1}}} \\
DK:\left\{ \begin{array}{l}
\dfrac{{{x^2} – 9}}{{x – 1}} \ge 0\\
x – 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 1\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} – 9 \ge 0\\
x – 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} – 9 \le 0\\
x – 1 < 0
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 1\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le – 3
\end{array} \right.\\
x > 1
\end{array} \right.\\
\left\{ \begin{array}{l}
– 3 \le x \le 3\\
x < 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 3\\
– 3 \le x < 1
\end{array} \right.
\end{array}\)