a) Phương trình \(1,5{x^2}-{\rm{ }}1,6x{\rm{ }} + {\rm{ }}0,1{\rm{ }} = {\rm{ }}0\)
Có \(a + b + c = 1,5 – 1,6 + 0,1 = 0\) nên \(\displaystyle{x_1} = 1;{x_2} = {\rm{ }}{{0,1} \over {1,5}} = {1 \over {15}}\)
b) Phương trình \(\sqrt 3 {x^2}-{\rm{ }}\left( {1{\rm{ }} - {\rm{ }}\sqrt 3 } \right)x{\rm{ }}-{\rm{ }}1{\rm{ }} = {\rm{ }}0\)
Có \(a – b + c = \sqrt{3} + (1 - \sqrt{3}) + (-1) = 0\) nên \(\displaystyle{x_1} = - 1,{x_2} = - {{ - 1} \over {\sqrt 3 }} = {\rm{ }}{{\sqrt 3 } \over 3}\)
c) \(\left( {2{\rm{ }} - {\rm{ }}\sqrt 3 } \right){x^2} + {\rm{ }}2\sqrt 3 x{\rm{ }}-{\rm{ }}\left( {2{\rm{ }} + {\rm{ }}\sqrt 3 } \right){\rm{ }} = {\rm{ }}0\)
Có \(a + b + c = 2 - \sqrt{3} + 2\sqrt{3} – (2 + \sqrt{3}) = 0\)
\({x_1} = 1;{x_2} = \dfrac{{ - \left( {2 + \sqrt 3 } \right)}}{{2 - \sqrt 3 }} = \dfrac{{ - {{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} = - 7 - 4\sqrt 3 \)
d) \(\left( {m{\rm{ }}-{\rm{ }}1} \right){x^2}-{\rm{ }}\left( {2m{\rm{ }} + {\rm{ }}3} \right)x{\rm{ }} + {\rm{ }}m{\rm{ }} + {\rm{ }}4{\rm{ }} = {\rm{ }}0\)
Có \(a + b + c = m – 1 – (2m + 3) + m + 4 = 0\)
Nên \(\displaystyle{x_1} = 1,{x_2} = {\rm{ }}{{m + 4} \over {m - 1}}\)
