\(a)\) \({x^2} - 3x + 1 = 0\)
\(\Leftrightarrow \displaystyle {x^2} - 2.{3 \over 2}x + {9 \over 4} = {9 \over 4} - 1\)
\( \Leftrightarrow \displaystyle {\left( {x - {3 \over 2}} \right)^2} = {5 \over 4} \)
\(\Leftrightarrow \displaystyle \left| {x - {3 \over 2}} \right| = {{\sqrt 5 } \over 2}\)
\( \Leftrightarrow \displaystyle x - {3 \over 2} = {{\sqrt 5 } \over 2}\) hoặc \(x -\displaystyle {3 \over 2} = - {{\sqrt 5 } \over 2}\)
\( \Leftrightarrow \displaystyle x = {{3 + \sqrt 5 } \over 2}\) hoặc \(x =\displaystyle {{3 - \sqrt 5 } \over 2}\)
Vậy phương trình có hai nghiệm: \({x_1} = \displaystyle {{3 + \sqrt 5 } \over 2};\)\({x_2} = \displaystyle{{3 - \sqrt 5 } \over 2}\)
\(b)\) \({x^2} + \sqrt 2 x - 1 = 0\)
\( \Leftrightarrow \displaystyle{x^2} + 2.{{\sqrt 2 } \over 2}x + {\left( {{{\sqrt 2 } \over 2}} \right)^2} \)\(= \displaystyle1 + {\left( {{{\sqrt 2 } \over 2}} \right)^2}\)
\( \Leftrightarrow\displaystyle {\left( {x + {{\sqrt 2 } \over 2}} \right)^2} = {3 \over 2} \)
\(\Leftrightarrow \displaystyle\left| {x + {{\sqrt 2 } \over 2}} \right| = {{\sqrt 6 } \over 2}\)
\( \Leftrightarrow\displaystyle x + {{\sqrt 2 } \over 2} = {{\sqrt 6 } \over 2}\) hoặc \(x + \displaystyle {{\sqrt 2 } \over 2} = - {{\sqrt 6 } \over 2}\)
\( \Leftrightarrow x =\displaystyle {{ - \sqrt 2 + \sqrt 6 } \over 2}\) hoặc \(x = \displaystyle - {{\sqrt 2 + \sqrt 6 } \over 2}\)
Vậy phương trình có hai nghiệm: \({x_1} = \displaystyle {{ - \sqrt 2 + \sqrt 6 } \over 2};\)\({x_2} =\displaystyle - {{\sqrt 2 + \sqrt 6 } \over 2}\)
\(c)\) \( 5{x^2} - 7x + 1 = 0 \)
\(\Leftrightarrow \displaystyle{x^2} - {7 \over 5}x + {1 \over 5} = 0 \)
\( \Leftrightarrow\displaystyle {x^2} - 2.{7 \over {10}}x + {{49} \over {100}} = {{49} \over {100}} - {1 \over 5} \)
\( \Leftrightarrow\displaystyle {\left( {x - {7 \over {10}}} \right)^2} = {{29} \over {100}} \)
\(\Leftrightarrow \displaystyle\left| {x - {7 \over {10}}} \right| = {{\sqrt {29} } \over {10}} \)
\( \Leftrightarrow \displaystyle x - {7 \over {10}} = {{\sqrt {29} } \over {10}}\) hoặc \(x - \displaystyle {7 \over {10}} = - {{\sqrt {29} } \over {10}}\)
\( \Leftrightarrow x = \displaystyle {{7 + \sqrt {29} } \over {10}}\) hoặc \(x = \displaystyle {{7 - \sqrt {29} } \over {10}}\)
Vậy phương trình có hai nghiệm: \({x_1} = \displaystyle {{7 + \sqrt {29} } \over {10}};\)\({x_2} =\displaystyle \displaystyle {{7 - \sqrt {29} } \over {10}}\)
\(d)\) \( 3{x^2} + 2\sqrt 3 x - 2 = 0 \)
\( \Leftrightarrow \displaystyle{x^2} + 2.{{\sqrt 3 } \over 3}x - {2 \over 3} = 0 \)
\( \Leftrightarrow \displaystyle x + 2.{{\sqrt 3 } \over 3}x + {\left( {{{\sqrt 3 } \over 3}} \right)^2}\)\( = \displaystyle{2 \over 3} + {\left( {{{\sqrt 3 } \over 3}} \right)^2} \)
\( \Leftrightarrow \displaystyle {\left( {x + {{\sqrt 3 } \over 3}} \right)^2} = 1 \)
\(\Leftrightarrow \displaystyle \left| {x + {{\sqrt 3 } \over 3}} \right| = 1 \)
\( \Leftrightarrow \displaystyle x + {{\sqrt 3 } \over 3} = 1\) hoặc \(x + \displaystyle {{\sqrt 3 } \over 3} = - 1\)
\( \Leftrightarrow \displaystyle x = 1 - {{\sqrt 3 } \over 3}\) hoặc \(x = - 1 - \displaystyle {{\sqrt 3 } \over 3}\)
Vậy phương trình có hai nghiệm: \({x_1} = 1 - \displaystyle{{\sqrt 3 } \over 3};\)\({x_2} = - 1 - \displaystyle {{\sqrt 3 } \over 3}\)
