Question

[hoc24]

1. Tìm x:

a) \(\left|3x-5\right|-\left|x+2\right|=0\)

b)\(\left|3x-5\right|+\left|x+2\right|=0\)

c)\(\left|3x-5\right|-x+2=0\)

d)\(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)

e)\(\dfrac{11}{2}\le\left|x+2\right|\le\dfrac{17}{2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: =>|3x-5|=|x+2|

=>3x-5=x+2 hoặc 3x-5=-x-2

=>2x=7 hoặc 4x=3

=>x=7/2 hoặc x=3/4

b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

c: \(\Leftrightarrow\left|3x-5\right|=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)