Giai các phương trình và bất phương trình sau : a) x+1/3 = x-3 b) 2/x – 4/1-x = 3x+4/x^2-x c) căn 2x+1 = 7-x 27/08/2021 Bởi Rylee Giai các phương trình và bất phương trình sau : a) x+1/3 = x-3 b) 2/x – 4/1-x = 3x+4/x^2-x c) căn 2x+1 = 7-x
Đáp án: \(\begin{array}{l}a)\,\,x = 5\\b)\,\,x = 2\\c)\,\,x = 4.\end{array}\) Giải thích các bước giải: \(\begin{array}{l}a)\,\,\frac{{x + 1}}{3} = x – 3 \Leftrightarrow x + 1 = 3\left( {x – 3} \right)\\ \Leftrightarrow x + 1 = 3x – 9\\ \Leftrightarrow 2x = 10\\ \Leftrightarrow x = 5.\\b)\,\,\frac{2}{x} – \frac{4}{{1 – x}} = \frac{{3x + 4}}{{{x^2} – x}}\\DK:\,\,\,x \ne 0,\,\,\,x \ne 1\\pt \Leftrightarrow \frac{2}{x} + \frac{4}{{x – 1}} = \frac{{3x + 4}}{{x\left( {x – 1} \right)}}\\ \Leftrightarrow 2\left( {x – 1} \right) + 4x = 3x + 4\\ \Leftrightarrow 2x – 2 + 4x = 3x + 4\\ \Leftrightarrow 3x = 6\\ \Leftrightarrow x = 2.\\c)\,\,\,\sqrt {2x + 1} = 7 – x\\DK:\,\,\,x \ge – \frac{1}{2}\\pt \Leftrightarrow \left\{ \begin{array}{l}7 – x \ge 0\\2x + 1 = {\left( {7 – x} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le 7\\2x + 1 = 49 – 14x + {x^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \le 7\\{x^2} – 16x + 48 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le 7\\\left[ \begin{array}{l}x = 4\\x = 12\end{array} \right.\end{array} \right. \Leftrightarrow x = 4.\end{array}\) Bình luận
Đáp án:
\(\begin{array}{l}
a)\,\,x = 5\\
b)\,\,x = 2\\
c)\,\,x = 4.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,\frac{{x + 1}}{3} = x – 3 \Leftrightarrow x + 1 = 3\left( {x – 3} \right)\\
\Leftrightarrow x + 1 = 3x – 9\\
\Leftrightarrow 2x = 10\\
\Leftrightarrow x = 5.\\
b)\,\,\frac{2}{x} – \frac{4}{{1 – x}} = \frac{{3x + 4}}{{{x^2} – x}}\\
DK:\,\,\,x \ne 0,\,\,\,x \ne 1\\
pt \Leftrightarrow \frac{2}{x} + \frac{4}{{x – 1}} = \frac{{3x + 4}}{{x\left( {x – 1} \right)}}\\
\Leftrightarrow 2\left( {x – 1} \right) + 4x = 3x + 4\\
\Leftrightarrow 2x – 2 + 4x = 3x + 4\\
\Leftrightarrow 3x = 6\\
\Leftrightarrow x = 2.\\
c)\,\,\,\sqrt {2x + 1} = 7 – x\\
DK:\,\,\,x \ge – \frac{1}{2}\\
pt \Leftrightarrow \left\{ \begin{array}{l}
7 – x \ge 0\\
2x + 1 = {\left( {7 – x} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
2x + 1 = 49 – 14x + {x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
{x^2} – 16x + 48 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
\left[ \begin{array}{l}
x = 4\\
x = 12
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 4.
\end{array}\)