Biết $\tan x = \dfrac{{2b}}{{a - c}}$. Giá trị của biểu thức $A = a{\cos ^2}x + 2b\sin x.cosx + c{\sin ^2}x$ bằng

$A = a{\cos ^2}x + 2b\sin x.cosx + c{\sin ^2}x$$ \Leftrightarrow \dfrac{A}{{{{\cos }^2}x}} = a + 2b\tan x + c{\tan ^2}x$

$ \Leftrightarrow A\left( {1 + {{\tan }^2}x} \right) = a + 2b\tan x + c{\tan ^2}x$$ \Leftrightarrow A\left( {1 + {{\left( {\dfrac{{2b}}{{a - c}}} \right)}^2}} \right) = a + 2b\dfrac{{2b}}{{a - c}} + c{\left( {\dfrac{{2b}}{{a - c}}} \right)^2}$

$ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}\left( {a - c} \right) + c4{b^2}}}{{{{\left( {a - c} \right)}^2}}}$

$ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}a}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a.\left( {{{\left( {a - c} \right)}^2} + 4{b^2}} \right)}}{{{{\left( {a - c} \right)}^2}}}$$ \Leftrightarrow A = a$.