Question

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

Answer 1

\(\frac{1}{16}=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)

=> \(\left(\frac{1}{3}-2x\right)=\frac{1}{4}\)hoặc \(\left(-\frac{1}{4}\right)\)

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

\(\Rightarrow2x=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\Rightarrow x=\frac{1}{12}:2=\frac{1}{24}\)

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

\(\Rightarrow2x=\frac{1}{3}-\left(-\frac{1}{4}\right)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\Rightarrow x=\frac{7}{12}:2=\frac{7}{24}\)

Vậy \(x\in\left\{\frac{1}{24};\frac{7}{24}\right\}\)