Giải các bất phương trình: a, 2x+1/6 -x > 3- x-2/9 b, x^2 -2x<0 c, 3x-2/x+4>0 03/08/2021 Bởi Sadie Giải các bất phương trình: a, 2x+1/6 -x > 3- x-2/9 b, x^2 -2x<0 c, 3x-2/x+4>0
Đáp án: c. \(\left[ \begin{array}{l}x > \dfrac{2}{3}\\x < – 4\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\dfrac{{2x + 1}}{6} – x > 3 – \dfrac{{x – 2}}{9}\\ \to \dfrac{{9\left( {2x + 1} \right) – 36x}}{{36}} > \dfrac{{3.36 – 6\left( {x – 2} \right)}}{{36}}\\ \to 9\left( {2x + 1} \right) – 36x > 3.36 – 6\left( {x – 2} \right)\\ \to 18x + 9 – 36x > 108 – 6x + 12\\ \to 12x < – 111\\ \to x < – \dfrac{{111}}{{12}}\\b.{x^2} – 2x < 0\\ \to x\left( {x – 2} \right) < 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 0\\x – 2 < 0\end{array} \right.\\\left\{ \begin{array}{l}x < 0\\x – 2 > 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 0\\x < 2\end{array} \right.\\\left\{ \begin{array}{l}x < 0\\x > 2\end{array} \right.\left( l \right)\end{array} \right.\\c.\dfrac{{3x – 2}}{{x + 4}} > 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}3x – 2 > 0\\x + 4 > 0\end{array} \right.\\\left\{ \begin{array}{l}3x – 2 < 0\\x + 4 < 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x > \dfrac{2}{3}\\x > – 4\end{array} \right.\\\left\{ \begin{array}{l}x < \dfrac{2}{3}\\x < – 4\end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}x > \dfrac{2}{3}\\x < – 4\end{array} \right.\end{array}\) Bình luận
Đáp án:
c. \(\left[ \begin{array}{l}
x > \dfrac{2}{3}\\
x < – 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{2x + 1}}{6} – x > 3 – \dfrac{{x – 2}}{9}\\
\to \dfrac{{9\left( {2x + 1} \right) – 36x}}{{36}} > \dfrac{{3.36 – 6\left( {x – 2} \right)}}{{36}}\\
\to 9\left( {2x + 1} \right) – 36x > 3.36 – 6\left( {x – 2} \right)\\
\to 18x + 9 – 36x > 108 – 6x + 12\\
\to 12x < – 111\\
\to x < – \dfrac{{111}}{{12}}\\
b.{x^2} – 2x < 0\\
\to x\left( {x – 2} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x – 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x – 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x > 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
c.\dfrac{{3x – 2}}{{x + 4}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x – 2 > 0\\
x + 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x – 2 < 0\\
x + 4 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \dfrac{2}{3}\\
x > – 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \dfrac{2}{3}\\
x < – 4
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{2}{3}\\
x < – 4
\end{array} \right.
\end{array}\)