Bài 1:giải các phương trình sau: a)|3x|=x+7 b)|-4x|=-2x+11 c)|x-9|=2x+5 d)|3x|-1=4x+1 e)|3x|-x-4 24/10/2021 Bởi Cora Bài 1:giải các phương trình sau: a)|3x|=x+7 b)|-4x|=-2x+11 c)|x-9|=2x+5 d)|3x|-1=4x+1 e)|3x|-x-4
a) `|3x|=x+7` ⇔\(\left[ \begin{array}{l}3x=x+7\\-3x=x+7\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=7\\x=-4x=7\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{7}{2}\\x=\frac{-4}{7}\end{array} \right.\) b) `|-4x|=-2x+11` ⇔\(\left[ \begin{array}{l}4x=-2x+11\\-4x=-2x+11\end{array} \right.\) ⇔\(\left[ \begin{array}{l}6x=11\\-2x=11\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\frac{11}{6}\\x=\frac{-11}{2}\end{array} \right.\) c) `|x-9|=2x+5` ⇔\(\left[ \begin{array}{l}x-9=2x+5\\9-x=2x+5\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-14\\-3x=-4\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-14\\x=\frac{4}{3}\end{array} \right.\) d) `|3x|-1=4x+1` ⇔`|3x|=4x+2` ⇔\(\left[ \begin{array}{l}3x=4x+2\\-3x=4x+2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-2\\7x=-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-2\\x=\frac{-2}{7}\end{array} \right.\) e) `|3x|-x-4 ` ⇔\(\left[ \begin{array}{l}3x-x-4\\-3x-x-4\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x-4\\-4x-4\end{array} \right.\) Bình luận
a)
`|3x|=x+7`
⇔\(\left[ \begin{array}{l}3x=x+7\\-3x=x+7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=7\\x=-4x=7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{7}{2}\\x=\frac{-4}{7}\end{array} \right.\)
b)
`|-4x|=-2x+11`
⇔\(\left[ \begin{array}{l}4x=-2x+11\\-4x=-2x+11\end{array} \right.\)
⇔\(\left[ \begin{array}{l}6x=11\\-2x=11\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{11}{6}\\x=\frac{-11}{2}\end{array} \right.\)
c)
`|x-9|=2x+5`
⇔\(\left[ \begin{array}{l}x-9=2x+5\\9-x=2x+5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-14\\-3x=-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-14\\x=\frac{4}{3}\end{array} \right.\)
d)
`|3x|-1=4x+1`
⇔`|3x|=4x+2`
⇔\(\left[ \begin{array}{l}3x=4x+2\\-3x=4x+2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\7x=-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=\frac{-2}{7}\end{array} \right.\)
e)
`|3x|-x-4 `
⇔\(\left[ \begin{array}{l}3x-x-4\\-3x-x-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x-4\\-4x-4\end{array} \right.\)