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在方程环境中使用 flushleft 时出错

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在方程环境中使用 flushleft 时出错

我在LaTeX中练习，遇到了这样的问题：

! Missing $ inserted.
<inserted text> 
                $
l.28        \begin{flushleft}

代码

\documentclass[a4paper,12pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T2A]{fontenc}
\usepackage{textcomp}
\usepackage[english,russian]{babel}
\usepackage{amsmath, amssymb}


\usepackage{geometry} % Меняем поля страницы
\geometry{left=2cm}% левое поле
\geometry{right=1.5cm}% правое поле
\geometry{top=1cm}% верхнее поле
\geometry{bottom=2cm}% нижнее поле

\begin{document}


  30.4 в \\
  $\sqrt{\frac{5x-1}{x+3}} =  2$ \\
  $\frac{5x-1}{x+3} = 4, x\neq -3$ \\
  $5x-1 = 4x+12$ \\
  $x = 11$ \\  
  Ответ: 11.\\ \\
  30.6 в \\

  $\sqrt{3x+4} = \sqrt{5x+2} $
  \begin{equation*}
    \begin{flushleft}
  \begin{cases}
      3x+4 \ge 0 \\
      5x+2 \ge 0
  \end{cases}
  \begin{cases}
    x\ge -1.25 \\ 
    x \ge  -0.4
  \end{cases}
  \rightarrow D(x) \in [-0.4; +\infty) \\
\end{flushleft}
\end{equation*}

  $3x+4=5x+2$ \\
  $2 = 2x$ \\
   $x = 1$ \\
   $x \in D(x)$ \\
   Ответ: 1. \\ \\
   30.11 в \\

   $\sqrt{15+3x} = 1-x,    x\le 1 \Rightarrow D(x) \in (-\infty; 1] $\\
  $15+3x = x^2-2x+1$ \\
  $x^2 - 5x - 14 = 0$ \\
  $D = 25+4 \cdot 14 = 25 + 56 = 81$ \\
  $x_1 = \frac{5+9}{2} = 7$ \\
  $x_2 = \frac{5-9}{2} = -2$ \\
  $x_1 \not\in D(x)$ \\
  Ответ: -2. \\ \\
  30.18 в \\
  $\sqrt{x^2+x+1} = x+2, x\ge -2 \rightarrow D(x) \in [-2; +\infty) $ \\
  $x^2+x+1 = x^2 + 4x + 4$ \\
  $3x = -3$ \\
   $x = -1$ \\
   Ответ: -1. \\ \\
   30.21 в \\
   $\sqrt{3x+1} + \sqrt{x-4} = 2\sqrt{x} $ 
   \begin{equation*}
     \begin{flushleft}
    \begin{cases}
        3x+1 \ge  0 \\
      x-4 \ge  0 \\
      x \ge  0 \\
    \end{cases}
    \begin{cases}
      x \ge  -\frac{1}{3} \\
      x \ge  4 \\
      x \ge  0\\
    \end{cases}
    \Rightarrow D(x) \in [4; +\infty) \\
  \end{flushleft}
   \end{equation*}
   $3x+1+x-4 +2\sqrt{(3x+1)(x-4)} = 4x$ \\
   $4x-3+2\sqrt{3x^2-11x -4 } = 4x $ \\
   $2\sqrt{3x^2-11x-4}=3 $ 
   $12x^2-44x-16=9$ \\
   $12x^2-44x-25 = 0$ \\
   $D_1 = 22^2 + 12 \cdot 25 = 484 +300 = 784$ \\
   $x_1 = \frac{22+28}{12} = \frac{25}{6} = 4\frac{1}{6}$ \\
   $x_2 = \frac{22-28}{12} = -\frac{1}{2}$ \\
   $x_2 \not\in D(x)$ \\
   Ответ: $4\frac{1}{6}$\\
\end{document}

我尝试在方程组中放入 $$，但没有用。这个错误的原因是什么？如何缩进解决方案，以便在视觉上将其与作业编号区分开来？

答案1

  1. 我还添加了T1 fontenc俄语语言或字符的重音（或特殊字符）的可能问题；

  2. 我已经添加了\usepackage[english,main=russian]{babel} https://ctan.math.illinois.edu/macros/latex/contrib/babel-contrib/russian/russianb.pdf声明你的俄语为主要文本（见第 4 页）。 在此处输入图片描述

  3. 我已将主要问题删除\begin{flushleft} ...\end{flushleft}到环境中，\begin{equation}....\end{equation}因为该flushleft环境用于将文本左对齐而不是对齐到等式中（介于两者之间的所有文本\begin{flushleft}都是\end{flushleft}左对齐的）。

\documentclass[a4paper,12pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T2A,T1]{fontenc}
\usepackage{textcomp}
\usepackage[english,main=russian]{babel}
\usepackage{amsmath, amssymb}


\usepackage{geometry} % Меняем поля страницы
\geometry{left=2cm}% левое поле
\geometry{right=1.5cm}% правое поле
\geometry{top=1cm}% верхнее поле
\geometry{bottom=2cm}% нижнее поле

\begin{document}


  30.4 в \\
  $\sqrt{\frac{5x-1}{x+3}} =  2$ \\
  $\frac{5x-1}{x+3} = 4, x\neq -3$ \\
  $5x-1 = 4x+12$ \\
  $x = 11$ \\  
  Ответ: 11.\\ \\
  30.6 в \\

  $\sqrt{3x+4} = \sqrt{5x+2} $
  \begin{equation*}
   \begin{cases}
      3x+4 \ge 0 \\
      5x+2 \ge 0
  \end{cases}
  \begin{cases}
    x\ge -1.25 \\ 
    x \ge  -0.4
  \end{cases}
  \rightarrow D(x) \in [-0.4; +\infty) \\
\end{equation*}

  $3x+4=5x+2$ \\
  $2 = 2x$ \\
   $x = 1$ \\
   $x \in D(x)$ \\
   Ответ: 1. \\ \\
   30.11 в \\

   $\sqrt{15+3x} = 1-x,    x\le 1 \Rightarrow D(x) \in (-\infty; 1] $\\
  $15+3x = x^2-2x+1$ \\
  $x^2 - 5x - 14 = 0$ \\
  $D = 25+4 \cdot 14 = 25 + 56 = 81$ \\
  $x_1 = \frac{5+9}{2} = 7$ \\
  $x_2 = \frac{5-9}{2} = -2$ \\
  $x_1 \not\in D(x)$ \\
  Ответ: -2. \\ \\
  30.18 в \\
  $\sqrt{x^2+x+1} = x+2, x\ge -2 \rightarrow D(x) \in [-2; +\infty) $ \\
  $x^2+x+1 = x^2 + 4x + 4$ \\
  $3x = -3$ \\
   $x = -1$ \\
   Ответ: -1. \\ \\
   30.21 в \\
   $\sqrt{3x+1} + \sqrt{x-4} = 2\sqrt{x} $ 
   \begin{equation*}
        \begin{cases}
        3x+1 \ge  0 \\
      x-4 \ge  0 \\
      x \ge  0 \\
    \end{cases}
    \begin{cases}
      x \ge  -\frac{1}{3} \\
      x \ge  4 \\
      x \ge  0\\
    \end{cases}
    \Rightarrow D(x) \in [4; +\infty) \\
    \end{equation*}
   $3x+1+x-4 +2\sqrt{(3x+1)(x-4)} = 4x$ \\
   $4x-3+2\sqrt{3x^2-11x -4 } = 4x $ \\
   $2\sqrt{3x^2-11x-4}=3 $ 
   $12x^2-44x-16=9$ \\
   $12x^2-44x-25 = 0$ \\
   $D_1 = 22^2 + 12 \cdot 25 = 484 +300 = 784$ \\
   $x_1 = \frac{22+28}{12} = \frac{25}{6} = 4\frac{1}{6}$ \\
   $x_2 = \frac{22-28}{12} = -\frac{1}{2}$ \\
   $x_2 \not\in D(x)$ \\
   Ответ: $4\frac{1}{6}$\\
\end{document}

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