ko biết đâu
\(x^2-\frac{7}{6}x+\frac{1}{3}=0\)
=> \(\left[x^2-2\cdot x\cdot\frac{7}{12}+\left(\frac{7}{12}\right)^2\right]-\frac{1}{144}=0\)
=> \(\left(x-\frac{7}{12}\right)^2-\frac{1}{144}=0\)
=> \(\left(x-\frac{7}{12}\right)^2-\left(\frac{1}{12}\right)^2=0\)
=> \(\left(x-\frac{7}{12}+\frac{1}{12}\right)\left(x-\frac{7}{12}-\frac{1}{12}\right)=0\)
=> \(\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)
=> x = 1/2 hoặc x = 2/3
Bài làm :
Ta có :
\(x^2-\frac{7}{6}x+\frac{1}{3}=0\)
\(\Leftrightarrow\left[x^2-2.x.\frac{7}{12}+\left(\frac{7}{12}\right)^2\right]-\frac{1}{144}=0\)
\(\Leftrightarrow\left(x-\frac{7}{12}\right)^2-\frac{1}{144}=0\)
\(\Leftrightarrow\left(x-\frac{7}{12}\right)^2-\left(\frac{1}{12}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{7}{12}+\frac{1}{12}\right)\left(x-\frac{7}{12}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\x-\frac{2}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
Vậy x = 1/2 hoặc x = 2/3
