Question

tim x, biet :

b,|\(\dfrac{x}{2}\)-1|=3

c, |-x+\(\dfrac{2}{5}\)|+\(\dfrac{1}{2}\)=3,5

Answer 1

|x/2-1|=3

=>x/2-1=3 hoặc -3

=>x/2= 4 hoặc -2

=>x:2=4 hoặc -2

=>x=2 hoặc -1

Answer 2

b,|$\frac{x}{2}$-1|=3 \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

c, |-x+$\frac{2}{5}$|+$\frac{1}{2}$=3,5 \(\Leftrightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\) \(\Leftrightarrow\left|-x+\dfrac{2}{5}\right|=3\) \(\Leftrightarrow\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\\x=\dfrac{-7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{-13}{5}\\-x=\dfrac{7}{2}\end{matrix}\right.\)

Answer 3

b , \(\left|\dfrac{x}{2}-1\right|=3\)

=> \(\dfrac{x}{2}-1=\pm3\)

=> \(\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.=>\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\-4\end{matrix}\right.\)

Vậy x=8 hoặc x= -4

c, \(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)

\(\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\)

\(\left|-x+\dfrac{2}{5}\right|=3\)

=> \(\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.=>\left[{}\begin{matrix}-x=\dfrac{-13}{5}\\-x=\dfrac{7}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{13}{5}\\x=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy x = \(\dfrac{13}{5}\) hoặc x = \(\dfrac{-7}{2}\)