Question

Tìm x: a,[-3x-1]=\(\dfrac{1}{2}\)

b,5[x-3]=\(\dfrac{5}{7}\)

c,[x-\(\dfrac{1}{4}\)]=\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)

Chú thích: Dấu [ và ] là giá trị tuyệt đối

Answer 1

Tìm x:

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

\(\Rightarrow-3x-1=\dfrac{1}{2}\)hoặc\(-3x-1=-\dfrac{1}{2}\)

\(.-3x-1=\dfrac{1}{2}\)

\(-3x=\dfrac{1}{2}+1\)

\(-3x=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}:\left(-3\right)\)

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

.\(-3x-1=-\dfrac{1}{2}\)

\(-3x=-\dfrac{1}{2}+1\)

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

\(x=\dfrac{1}{2}:\left(-3\right)\)

\(x=-\dfrac{1}{6}\)

Answer 2

Gíup mình đi nhé!

Answer 3

b, \(5.\left|x-3\right|=\dfrac{5}{7}\)

\(\left|x-3\right|=\dfrac{5}{7}:5\)

\(\left|x-3\right|=\dfrac{1}{7}\)

\(\Rightarrow x-3=\dfrac{1}{7}\) hoặc \(x-3=-\dfrac{1}{7}\)

. \(x-3=\dfrac{1}{7}\)

\(x=\dfrac{1}{7}+3\)

\(x=\dfrac{22}{7}\)

. \(x-3=-\dfrac{1}{7}\)

\(x=-\dfrac{1}{7}+3\)

\(x=\dfrac{20}{7}\)

Answer 4

c,\(\left|x-\dfrac{1}{4}\right|=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\left|x-\dfrac{1}{4}\right|=\dfrac{1}{12}\)

\(\Rightarrow x-\dfrac{1}{4}=\dfrac{1}{12}\) hoặc \(x-\dfrac{1}{4}=-\dfrac{1}{12}\)

. \(x-\dfrac{1}{4}=\dfrac{1}{12}\)

\(x=\dfrac{1}{12}+\dfrac{1}{4}\)

\(x=\dfrac{1}{3}\)

.\(x-\dfrac{1}{4}=-\dfrac{1}{12}\)

\(x=-\dfrac{1}{12}+\dfrac{1}{4}\)

\(x=\dfrac{1}{6}\)