Ответ:
Правило :
если [tex]\bf |\, x\, |\leq a[/tex] , то [tex]\bf -a\leq x\leq a[/tex] .
[tex]\bf |x-1|\leq 2x+1\ \ \ \Rightarrow \ \ \ -(2x+1)\leq x-1\leq 2x+1\ \ ,\\\\\left\{\begin{array}{l}\bf x-1\leq 2x+1\\\bf x-1\geq -2x-1\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2x\leq 1+1\\\bf x+2x\geq 1-1\end{array}\right\ \ \left\{\begin{array}{l}\bf -x\leq 2\\\bf 3x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\bf x\geq -2\\\bf x\geq 0\end{array}\right\ \ \Rightarrow \\\\\\x\geq 0\\\\Otvet:\ \ \boldsymbol{x\in [\ 0\ ;+\infty \, )\ .}[/tex]
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Ответ:
Правило :
если [tex]\bf |\, x\, |\leq a[/tex] , то [tex]\bf -a\leq x\leq a[/tex] .
[tex]\bf |x-1|\leq 2x+1\ \ \ \Rightarrow \ \ \ -(2x+1)\leq x-1\leq 2x+1\ \ ,\\\\\left\{\begin{array}{l}\bf x-1\leq 2x+1\\\bf x-1\geq -2x-1\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2x\leq 1+1\\\bf x+2x\geq 1-1\end{array}\right\ \ \left\{\begin{array}{l}\bf -x\leq 2\\\bf 3x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\bf x\geq -2\\\bf x\geq 0\end{array}\right\ \ \Rightarrow \\\\\\x\geq 0\\\\Otvet:\ \ \boldsymbol{x\in [\ 0\ ;+\infty \, )\ .}[/tex]