a)|3x|=x+7 b)|-4x|=-2x+11 c)|3-2x|=3x-7 20/08/2021 Bởi Melody a)|3x|=x+7 b)|-4x|=-2x+11 c)|3-2x|=3x-7
CHÚC BẠN HỌC TỐT!!! Giải thích các bước giải: $a, |3x|=x+7$ $(x\geq-7)$ \(⇔\left[ \begin{array}{l}3x=x+7\\3x=-x-7\end{array} \right.\) \(⇔\left[ \begin{array}{l}2x=7\\4x=-7\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{7}{4}\end{array} \right.\) Vậy $S=\{\dfrac{7}{2};-\dfrac{7}{4}\}$ $b, |-4x|=-2x+11$ $(x \leq \dfrac{11}{2})$ \(⇔\left[ \begin{array}{l}-4x=-2x+11\\-4x=2x-11\end{array} \right.\) \(⇔\left[ \begin{array}{l}2x=-11\\6x=11\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-\dfrac{11}{2}\\x=\dfrac{11}{6}\end{array} \right.\) Vậy $S=\{-\dfrac{11}{2};\dfrac{11}{6}\}$ $c, |3-2x|=3x-7$ $(x\geq\dfrac{7}{3})$ \(⇔\left[ \begin{array}{l}3-2x=3x-7\\3-2x=-3x+7\end{array} \right.\) \(⇔\left[ \begin{array}{l}5x=-10\\x=4\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-2(L)\\x=4\end{array} \right.\) Vậy $S=\{4\}$ Bình luận
CHÚC BẠN HỌC TỐT!!!
Giải thích các bước giải:
$a, |3x|=x+7$ $(x\geq-7)$
\(⇔\left[ \begin{array}{l}3x=x+7\\3x=-x-7\end{array} \right.\) \(⇔\left[ \begin{array}{l}2x=7\\4x=-7\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{7}{4}\end{array} \right.\)
Vậy $S=\{\dfrac{7}{2};-\dfrac{7}{4}\}$
$b, |-4x|=-2x+11$ $(x \leq \dfrac{11}{2})$
\(⇔\left[ \begin{array}{l}-4x=-2x+11\\-4x=2x-11\end{array} \right.\) \(⇔\left[ \begin{array}{l}2x=-11\\6x=11\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-\dfrac{11}{2}\\x=\dfrac{11}{6}\end{array} \right.\)
Vậy $S=\{-\dfrac{11}{2};\dfrac{11}{6}\}$
$c, |3-2x|=3x-7$ $(x\geq\dfrac{7}{3})$
\(⇔\left[ \begin{array}{l}3-2x=3x-7\\3-2x=-3x+7\end{array} \right.\) \(⇔\left[ \begin{array}{l}5x=-10\\x=4\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-2(L)\\x=4\end{array} \right.\)
Vậy $S=\{4\}$