giải pt `3x^2-7x+1=0`

Đáp án:

$x = \dfrac{7}{6}\pm\dfrac{\sqrt{37}}{6}$

Giải thích các bước giải:

$\begin{array}{l}\quad 3x^2 -7x + 1 =0\\ \Leftrightarrow 3\left(x^2 - 2\cdot \dfrac{7}{6}x + \dfrac{49}{36}\right)-\dfrac{37}{12} =0\\ \Leftrightarrow 3\left(x- \dfrac76\right)^2 = \dfrac{37}{12}\\ \Leftrightarrow \left(x- \dfrac76\right)^2 = \dfrac{37}{36}\\ \Leftrightarrow \left|x - \dfrac76\right| = \dfrac{\sqrt{37}}{6}\\ \Leftrightarrow \left[ \begin{array}{l}x - \dfrac76 = \dfrac{\sqrt{37}}{6}\\x - \dfrac76 = -\dfrac{\sqrt{37}}{6}\end{array}\right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac76+ \dfrac{\sqrt{37}}{6}\\x = \dfrac76 -\dfrac{\sqrt{37}}{6}\end{array}\right.\\ Vậy\,\,x = \dfrac{7}{6}\pm\dfrac{\sqrt{37}}{6} \end{array}$