Phương trình \(2\sqrt 3 {x^2} + x + 1 = \sqrt 3 \left( {x + 1} \right)\) có nghiệm

* Hướng dẫn giải

\(2\sqrt 3 {x^2} + x + 1 = \sqrt 3 \left( {x + 1} \right)\)

\(\begin{array}{l} \Leftrightarrow 2\sqrt 3 {x^2} + x + 1 - \sqrt 3 \left( {x + 1} \right) = 0\\ \Leftrightarrow 2\sqrt 3 {x^2} + x + 1 - \sqrt 3 x - \sqrt 3  = 0\\ \Leftrightarrow 2\sqrt 3 {x^2} + \left( {1 - \sqrt 3 } \right)x + 1 - \sqrt 3  = 0\end{array}\)

\(\Delta  = {\left( {1 - \sqrt 3 } \right)^2} - 4.2\sqrt 3 \left( {1 - \sqrt 3 } \right) \)\(= 4 - 2\sqrt 3  - 8\sqrt 3  + 24\)\( = 28 - 10\sqrt 3 \)\( = 25 - 2.5.\sqrt 3  + 3 \)\(= {\left( {5 - \sqrt 3 } \right)^2}\)\( \Rightarrow \sqrt \Delta   = 5 - \sqrt 3 \) 

\({x_1} = \dfrac{{\sqrt 3  - 1 + 5 - \sqrt 3 }}{{4\sqrt 3 }} \)\(= \dfrac{{\sqrt 3 }}{3};\)\({x_2} = \dfrac{{\sqrt 3  - 1 - 5 + \sqrt 3 }}{{4\sqrt 3 }} \)\(= \dfrac{{1 - \sqrt 3 }}{2}\)