Question

bài 1 tìm x biết 

a,\(|2x-3|-\frac{1}{3}=\text{0}\)  

b,\(\frac{5}{6}-|x+\frac{1}{4}|=\frac{1}{4}\)  

c,\(3-|2x+1,5|=\frac{5}{4}\) 

giúp mik với

 

Answer 1

a) \(\left|2x-3\right|-\frac{1}{3}=0\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{10}{3}\\2x=\frac{8}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

b) \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{7}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}\)

c) \(3-\left|2x+1,5\right|=\frac{5}{4}\)

\(\Leftrightarrow\left|2x+\frac{3}{2}\right|=\frac{7}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{2}=\frac{7}{4}\\2x+\frac{3}{2}=-\frac{7}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{13}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{13}{8}\end{cases}}\)

Answer 2

a. \(\left|2x-3\right|-\frac{1}{3}=0\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

b. \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{7}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}\)

c. \(3-\left|2x+1,5\right|=\frac{5}{4}\)

\(\Leftrightarrow\left|2x+\frac{3}{2}\right|=\frac{7}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{2}=\frac{7}{4}\\2x+\frac{3}{2}=-\frac{7}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{13}{8}\end{cases}}\)

Answer 3

a) \(\left|2x-3\right|-\frac{1}{3}=0\)

=> \(\left|2x-3\right|=\frac{1}{3}\Rightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{10}{3}\\2x=\frac{8}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{3};\frac{4}{3}\right\}\)

b) \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

=> \(\left|x+\frac{1}{4}\right|=\frac{7}{12}\)

=> \(\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}\)

=> \(x\in\left\{\frac{1}{3};-\frac{5}{6}\right\}\)

c) \(3-\left|2x+1,5\right|=\frac{5}{4}\)

=> \(\left|2x+1,5\right|=1,75\Rightarrow\orbr{\begin{cases}2x+1,5=1,75\\2x+1,5=-1,75\end{cases}}\Rightarrow\orbr{\begin{cases}2x=0,25\\2x=-3,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=0,125\\x=-1,625\end{cases}}\)

Vậy \(x\in\left\{0,125;-1,625\right\}\)

Answer 4

a ) \(\left|2x-3\right|-\frac{1}{3}=0\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{3}\\2x-3=-\frac{1}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)

b ) \(\frac{5}{6}-\left|x+\frac{1}{4}\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{7}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{7}{12}\\x+\frac{1}{4}=-\frac{7}{12}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{6}\end{cases}}}\)

c ) \(3-\left|2x+1,5\right|=\frac{5}{4}\)

\(\Leftrightarrow\left|2x+\frac{3}{2}\right|=\frac{7}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{2}=\frac{7}{4}\\2x+\frac{3}{2}=-\frac{7}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=-\frac{13}{8}\end{cases}}}\)