Biết \(\tan x = \dfrac{{2b}}{{a - c}}\). Giá trị c...

B

- Hướng dẫn giải

Phương pháp giải:

Chia cả 2 vế cho \({\cos ^2}x\).

Giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,A = a{\cos ^2}x + 2b\sin x\cos x + c{\sin ^2}x\\ \Leftrightarrow \dfrac{A}{{{{\cos }^2}x}} = a + 2b\dfrac{{\sin x}}{{\cos x}} + c\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\ \Leftrightarrow A\left( {1 + {{\tan }^2}x} \right) = a + 2b\tan x + c{\tan ^2}x\\ \Leftrightarrow A\left( {1 + {{\left( {\dfrac{{2b}}{{a - c}}} \right)}^2}} \right) = a + 2b\dfrac{{2b}}{{a - c}} + c{\left( {\dfrac{{2b}}{{a - c}}} \right)^2}\\ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}\left( {a - c} \right) + c.4{b^2}}}{{{{\left( {a - c} \right)}^2}}}\\ \Leftrightarrow A\dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}}}{{{{\left( {a - c} \right)}^2}}} = \dfrac{{a{{\left( {a - c} \right)}^2} + 4{b^2}a}}{{{{\left( {a - c} \right)}^2}}}\\ \Leftrightarrow A\left[ {{{\left( {a - c} \right)}^2} + {{\left( {2b} \right)}^2}} \right] = a\left[ {{{\left( {a - c} \right)}^2} + 4{b^2}} \right]\\ \Leftrightarrow A = a\end{array}\)

Chọn B