Bài 1
a)|x-1,7|=2,3
b) |x+3/4|-1/3=0
Bài 2
a) |2,5-x|=1,3
b) 1,6-|x-0,2|=0
c) |x-1,5|+ |2,5-x|=0
Bài 3
a) (1/7x-2/7).(-1/5x+3/5).(1/3x+4/3=0
b) 1/6x+1/10x-4/15x+1=0
Bài 1 a)|x-1,7|=2,3 b) |x+3/4|-1/3=0 Bài 2 a) |2,5-x|=1,3 b) 1,6-|x-0,2|=0 c) |x-1,5|+ |2,5-x|=0 Bài 3 a) (1/7x-2/7).(-1/5x+3/5).(1/3x+4/3=0 b) 1/6x+
By Alaia
Đáp án:
Giải thích các bước giải:
\(1/\\ a,\ |x-1,7|=2,3\\ ⇔\left[ \begin{array}{l}x-1,7=2,3\\x-1,7=-2,3\end{array} \right.\ ⇔\left[ \begin{array}{l}x=4\\x=-0,6\end{array} \right.\\ \text{Vậy x = 4 hoặc x = -0,6}\\ b,\ \bigg|x+\dfrac{3}{4}\bigg|-\dfrac{1}{3}=0\\ ⇔\bigg|x+\dfrac{3}{4}\bigg|=\dfrac{1}{3}\\ ⇔\left[ \begin{array}{l}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=\dfrac{-1}{3}\end{array} \right.\ ⇔\left[ \begin{array}{l}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{array} \right.\\ \text{Vậy x = $\dfrac{-5}{12}$ hoặc x = $\dfrac{-13}{12}$}\\ 2/\\ a,\ |2,5-x|=1,3\\ ⇔\left[ \begin{array}{l}2,5-x=1,3\\2,5-x=-1,3\end{array} \right.\ ⇔\left[ \begin{array}{l}x=1,2\\x=3,8\end{array} \right.\\ \text{Vậy x = 1,2 hoặc x = 3,8}\\ b,\ 1,6-|x-0,2|=0\\ ⇔|x-0,2|=1,6\\ ⇔\left[ \begin{array}{l}x-0,2=1,6\\x-0,2=-1,6\end{array} \right.\ ⇔\left[ \begin{array}{l}x=1,8\\x=-1,4\end{array} \right.\\ \text{Vậy x = 1,8 hoặc x = -1,4}\\ c,\ |x-1,5|+|2,5-x|=0\\ ⇔\left\{\begin{array}{l}x-1,5=0\\2,5-x=0\end{array} \right.⇔\left\{\begin{array}{l}x=1,5\\x=2,5\end{array} \right.\\ \text{Vậy x = 1,5 và x = 2,5}\\ 3/\\ a,\ \bigg(\dfrac{1}{7}x-\dfrac{2}{7} \bigg).\bigg(\dfrac{-1}{5}x+\dfrac{3}{5}\bigg).\bigg(\dfrac{1}{3}x+\dfrac{4}{3}\bigg)=0\\ ⇔\left[ \begin{array}{l}\dfrac{1}{7}x-\dfrac{2}{7}=0\\\dfrac{-1}{5}x+\dfrac{3}{5}=0\\\dfrac{1}{3}x+\dfrac{4}{3}=0\end{array} \right.\ ⇔\left[ \begin{array}{l}x=2\\x=3\\x=-4\end{array} \right.\\ \text{Vậy x = 2 hoặc x = 3 hoặc x = (-4)}\\ b, \dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\\ ⇔0x=(-1)\ \text{(vô lí)}\\ \text{Vậy x không tồn tại}\)
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