Trang chủ » Hỏi đáp » Môn Toán |x+1/3|-|-2/9|=0 2.|x|-1=(-1)^2021 |3x+1/2|-2/3=1 |x-2|+1/3=1 gấp lắm ạ 18/11/2024 |x+1/3|-|-2/9|=0 2.|x|-1=(-1)^2021 |3x+1/2|-2/3=1 |x-2|+1/3=1 gấp lắm ạ
Giải đáp: $\begin{array}{l}a)x = – \dfrac{5}{9};x = – \dfrac{1}{9}\\b)x = 0\\c)x = – \dfrac{{13}}{{18}};x = \dfrac{7}{{18}}\\d)x = \dfrac{4}{3};x = \dfrac{8}{3}\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a)\left| {x + \dfrac{1}{3}} \right| – \left| { – \dfrac{2}{9}} \right| = 0\\ \Leftrightarrow \left| {x + \dfrac{1}{3}} \right| – \dfrac{2}{9} = 0\\ \Leftrightarrow \left| {x + \dfrac{1}{3}} \right| = \dfrac{2}{9}\\ \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{1}{3} = \dfrac{2}{9}\\x + \dfrac{1}{3} = – \dfrac{2}{9}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{2}{9} – \dfrac{1}{3} = – \dfrac{1}{9}\\x = – \dfrac{2}{9} – \dfrac{1}{3} = – \dfrac{5}{9}\end{array} \right.\\Vậy\,x = – \dfrac{5}{9};x = – \dfrac{1}{9}\\b)2\left| x \right| – 1 = {\left( { – 1} \right)^{2021}}\\ \Leftrightarrow 2\left| x \right| – 1 = – 1\\ \Leftrightarrow 2\left| x \right| = 0\\ \Leftrightarrow \left| x \right| = 0\\ \Leftrightarrow x = 0\\Vậy\,x = 0\\c)\left| {3x + \dfrac{1}{2}} \right| – \dfrac{2}{3} = 1\\ \Leftrightarrow \left| {3x + \dfrac{1}{2}} \right| = 1 + \dfrac{2}{3}\\ \Leftrightarrow \left| {3x + \dfrac{1}{2}} \right| = \dfrac{5}{3}\\ \Leftrightarrow \left[ \begin{array}{l}3x + \dfrac{1}{2} = \dfrac{5}{3}\\3x + \dfrac{1}{2} = – \dfrac{5}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}3x = \dfrac{5}{3} – \dfrac{1}{2} = \dfrac{7}{6}\\3x = – \dfrac{5}{3} – \dfrac{1}{2} = \dfrac{{ – 13}}{6}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{7}{{18}}\\x = – \dfrac{{13}}{{18}}\end{array} \right.\\Vậy\,x = – \dfrac{{13}}{{18}};x = \dfrac{7}{{18}}\\d)\left| {x – 2} \right| + \dfrac{1}{3} = 1\\ \Leftrightarrow \left| {x – 2} \right| = 1 – \dfrac{1}{3} = \dfrac{2}{3}\\ \Leftrightarrow \left[ \begin{array}{l}x – 2 = \dfrac{2}{3}\\x – 2 = – \dfrac{2}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{2}{3} + 2 = \dfrac{8}{3}\\x = – \dfrac{2}{3} + 2 = \dfrac{4}{3}\end{array} \right.\\Vậy\,x = \dfrac{4}{3};x = \dfrac{8}{3}\end{array}$ Trả lời
a)x = – \dfrac{5}{9};x = – \dfrac{1}{9}\\
b)x = 0\\
c)x = – \dfrac{{13}}{{18}};x = \dfrac{7}{{18}}\\
d)x = \dfrac{4}{3};x = \dfrac{8}{3}
\end{array}$
a)\left| {x + \dfrac{1}{3}} \right| – \left| { – \dfrac{2}{9}} \right| = 0\\
\Leftrightarrow \left| {x + \dfrac{1}{3}} \right| – \dfrac{2}{9} = 0\\
\Leftrightarrow \left| {x + \dfrac{1}{3}} \right| = \dfrac{2}{9}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{3} = \dfrac{2}{9}\\
x + \dfrac{1}{3} = – \dfrac{2}{9}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{9} – \dfrac{1}{3} = – \dfrac{1}{9}\\
x = – \dfrac{2}{9} – \dfrac{1}{3} = – \dfrac{5}{9}
\end{array} \right.\\
Vậy\,x = – \dfrac{5}{9};x = – \dfrac{1}{9}\\
b)2\left| x \right| – 1 = {\left( { – 1} \right)^{2021}}\\
\Leftrightarrow 2\left| x \right| – 1 = – 1\\
\Leftrightarrow 2\left| x \right| = 0\\
\Leftrightarrow \left| x \right| = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
c)\left| {3x + \dfrac{1}{2}} \right| – \dfrac{2}{3} = 1\\
\Leftrightarrow \left| {3x + \dfrac{1}{2}} \right| = 1 + \dfrac{2}{3}\\
\Leftrightarrow \left| {3x + \dfrac{1}{2}} \right| = \dfrac{5}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{1}{2} = \dfrac{5}{3}\\
3x + \dfrac{1}{2} = – \dfrac{5}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{5}{3} – \dfrac{1}{2} = \dfrac{7}{6}\\
3x = – \dfrac{5}{3} – \dfrac{1}{2} = \dfrac{{ – 13}}{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{{18}}\\
x = – \dfrac{{13}}{{18}}
\end{array} \right.\\
Vậy\,x = – \dfrac{{13}}{{18}};x = \dfrac{7}{{18}}\\
d)\left| {x – 2} \right| + \dfrac{1}{3} = 1\\
\Leftrightarrow \left| {x – 2} \right| = 1 – \dfrac{1}{3} = \dfrac{2}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = \dfrac{2}{3}\\
x – 2 = – \dfrac{2}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3} + 2 = \dfrac{8}{3}\\
x = – \dfrac{2}{3} + 2 = \dfrac{4}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{4}{3};x = \dfrac{8}{3}
\end{array}$