Toán Lớp 7: Bài 1: Tìm x bt
a, | x – 1| = 3x – 2
b, |x – 3 | – x = 5
c, | x – 1/3 | + 4/5 = |(-3,2) + 2/5 |
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Toán Lớp 7: Bài 1: Tìm x bt
a, | x – 1| = 3x – 2
b, |x – 3 | – x = 5
c, | x – 1/3 | + 4/5 = |(-3,2) + 2/5 |
Comments ( 1 )
a)\left| {x – 1} \right| = 3x – 2\\
Dk:x \ge \dfrac{2}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 3x – 2\\
x – 1 = 2 – 3x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
4x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( {ktm} \right)\\
x = \dfrac{3}{4}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{3}{4}\\
b)\left| {x – 3} \right| – x = 5\\
\Leftrightarrow \left| {x – 3} \right| = x + 5\left( {dk:x \ge – 5} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x – 3 = x + 5\\
x – 3 = – x – 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– 3 = 5\left( {ktm} \right)\\
2x = – 2
\end{array} \right.\\
\Leftrightarrow x = – 1\left( {tm} \right)\\
Vậy\,x = – 1\\
c)\left| {x – \dfrac{1}{3}} \right| + \dfrac{4}{5} = \left| { – 3,2 + \dfrac{2}{5}} \right|\\
\Leftrightarrow \left| {x – \dfrac{1}{3}} \right| = \dfrac{{ – 4}}{5} + \left| {\dfrac{{ – 16}}{5} + \dfrac{2}{5}} \right|\\
\Leftrightarrow \left| {x – \dfrac{1}{3}} \right| = – \dfrac{4}{5} + \left| { – \dfrac{{14}}{5}} \right|\\
\Leftrightarrow \left| {x – \dfrac{1}{3}} \right| = – \dfrac{4}{5} + \dfrac{{14}}{5} = 2\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{1}{3} = 2\\
x – \dfrac{1}{3} = – 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2 + \dfrac{1}{3} = \dfrac{7}{3}\\
x = – 2 + \dfrac{1}{3} = \dfrac{{ – 5}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{3};x = – \dfrac{5}{3}
\end{array}$