2. Tìm x
a) \(5-\left|x+\dfrac{1}{2}\right|=1\)
\(\left|x+\dfrac{1}{2}\right|=5-1\)
\(\left|x+\dfrac{1}{2}\right|=4\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{-9}{2}\end{matrix}\right.\)
Vậy x=7/2 hoặc -9/2
b) \(\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)
\(\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)
\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=\dfrac{-17}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=\dfrac{-11}{3}\\\dfrac{1}{2}x=\dfrac{23}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-22}{3}\\x=\dfrac{46}{3}\end{matrix}\right.\)
Vậy x=-22/3 hoặc x=46/4
1. So sánh:
a. \(\dfrac{-1}{5}\) và \(\dfrac{1}{1000}\)
Ta có:
\(\dfrac{-1}{5}< 0\\ \dfrac{1}{1000}>0\\ \Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b.\(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)
Ta có:
\(\dfrac{267}{-268}>-1\\ \dfrac{-1347}{1343}< -1\\ \Rightarrow\dfrac{267}{-286}>\dfrac{-1347}{1343}\)
c.\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
Ta có:
\(\dfrac{-13}{38}< \dfrac{-13}{39}=\dfrac{-1}{3}\\ \dfrac{-29}{88}>\dfrac{-29}{87}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{-13}{38}< \dfrac{-29}{88}\)
1, So sánh
\(a,-\dfrac{1}{5}\) và \(\dfrac{1}{1000}\)
Cách 1: \(-\dfrac{1}{5}=-\dfrac{1.200}{5.200}=-\dfrac{200}{1000}\)
Ta có: \(-\dfrac{200}{1000}< \dfrac{1}{1000}\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)
Cách 2: Ta có:
\(-\dfrac{1}{5}\) là số âm, \(\dfrac{1}{1000}\) là số dương
âm < dương \(\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)
\(b,\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)
Ta có: \(\dfrac{267}{-268}< -1\left(1\right)\), \(\dfrac{-1347}{1343}>-1\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow\dfrac{267}{-268}>\dfrac{-1347}{1343}\)
\(c,\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
Ta có: \(\dfrac{-13}{38}=-\dfrac{13.44}{38.44}=-\dfrac{572}{1672}\)
\(\dfrac{29}{-88}=-\dfrac{29.19}{88.19}=-\dfrac{551}{1672}\)
\(\Rightarrow-\dfrac{572}{1672}< -\dfrac{551}{1672}\) hay \(-\dfrac{13}{38}< \dfrac{29}{-88}\)
Bài 2: Tìm x:
\(a,5-\left|x+\dfrac{1}{2}\right|=1\)
\(\Rightarrow\left|x+\dfrac{1}{2}\right|=4\)
Ta có \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{9}{2};\dfrac{7}{2}\right\}\)
\(b,\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)
\(\Rightarrow\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)
\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)
Ta có: \(\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(2-\dfrac{17}{3}\right):\dfrac{1}{2}=-\dfrac{22}{3}\\x=\left[2-\left(-\dfrac{17}{3}\right)\right]:\dfrac{1}{2}=\dfrac{46}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{22}{3};\dfrac{46}{3}\right\}\)
