Question

Tìm x

\((2x-\frac{1}{3})^2=\frac{1}{16}\)

CÁc bn giúp mik với mik đi học rồi

Answer 1

(2x-1/3)^2 = 1/16

=> 2x-1/3 = 1/4 hoặc 2x-1/3=-1/4

=> x=7/24 hoặc 1/24

Vậy .............

Tk mk nha

Answer 2

\(\left(2x-\frac{1}{3}\right)^2=\frac{1}{16}\)

\(\left(2x-\frac{1}{3}\right)^2=\left(\frac{1}{4}\right)^2hay\left(-\frac{1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{1}{4}\Rightarrow2x=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}\Rightarrow x=\frac{7}{24}\\2x-\frac{1}{3}=-\frac{1}{4}\Rightarrow2x=-\frac{1}{4}+\frac{1}{3}=\frac{1}{12}\Rightarrow x=\frac{1}{24}\end{cases}}\)

Vậy ..

học tốt 
 

Answer 3

ta có  \(\left(2x-\frac{1}{3}\right)^2=\frac{1}{16}\)

\(TH1\Rightarrow\left(2x-\frac{1}{3}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Leftrightarrow2x-\frac{1}{3}=\frac{1}{4}\Leftrightarrow2x=\frac{1}{4}+\frac{1}{3}\)

\(\Leftrightarrow2x=\frac{7}{12}\Rightarrow x=\frac{7}{12}:2=\frac{7}{24}\)

\(TH2\Rightarrow2x-\frac{1}{3}=-\frac{1}{4}\)

\(\Leftrightarrow2x=-\frac{1}{4}+\frac{1}{3}\)

\(\Leftrightarrow2x=\frac{1}{12}\Rightarrow x=\frac{1}{24}\)

Vậy \(x\in\left\{\frac{7}{24};\frac{1}{24}\right\}\)thì thỏa mãn đề bài