Cho các số a , b,c ,A ,B,C thoat mãn aC-2bB+cA=0 và ac-b^2>0 CMR AC-B^2<=0

Đáp án:

Giải thích các bước giải:

Ta có: \(aC - 2bB + cA = 0\) \( \Leftrightarrow aC + cA = 2bB\)

\( \Rightarrow {\left( {aC + cA} \right)^2} = 4{b^2}{B^2}\) \( \Leftrightarrow {a^2}{C^2} + {c^2}{A^2} + 2acCA = 4{b^2}{B^2}\)

\( \Leftrightarrow 2acCA = 4{b^2}{B^2} - {a^2}{C^2} - {c^2}{A^2}\) \( \Leftrightarrow AC = \dfrac{{4{b^2}{B^2} - {a^2}{C^2} - {c^2}{A^2}}}{{2ac}}\)

(do \(ac - {b^2} > 0\) nên \(ac > {b^2} \ge 0\) hay \(ac > 0\))

Khi đó \(AC - {B^2} = \dfrac{{4{b^2}{B^2} - {a^2}{C^2} - {c^2}{A^2}}}{{2ac}} - {B^2}\)

\( = \dfrac{{4{b^2}{B^2} - {a^2}{C^2} - {c^2}{A^2} - 2{B^2}ac}}{{2ac}}\) \( = \dfrac{{4{b^2}{B^2} - 4{B^2}ac - \left( {{a^2}{C^2} + {c^2}{A^2} - 2{B^2}ac} \right)}}{{2ac}}\)

\( = \dfrac{{4{B^2}\left( {{b^2} - ac} \right) - \left( {{a^2}{C^2} + {c^2}{A^2} - 2acCA + 2acCA - 2{B^2}ac} \right)}}{{2ac}}\)

\( = \dfrac{{4{B^2}\left( {{b^2} - ac} \right) - \left[ {{{\left( {aC - cA} \right)}^2} + 2ac\left( {CA - {B^2}} \right)} \right]}}{{2ac}}\)

\( = \dfrac{{4{B^2}\left( {{b^2} - ac} \right) - {{\left( {aC - cA} \right)}^2} - 2ac\left( {AC - {B^2}} \right)}}{{2ac}}\)

Mà \({b^2} - ac < 0,{\left( {aC - cA} \right)^2} \ge 0\) nên

\(\dfrac{{4{B^2}\left( {{b^2} - ac} \right) - {{\left( {aC - cA} \right)}^2} - 2ac\left( {AC - {B^2}} \right)}}{{2ac}}\)\( \le \dfrac{{0 - 0 - 2ac\left( {AC - {B^2}} \right)}}{{2ac}} = - AC + {B^2}\)

Hay \(AC - {B^2} \le - AC + {B^2}\) \( \Leftrightarrow 2AC - 2{B^2} \le 0\) \( \Leftrightarrow AC - {B^2} \le 0\)

Vậy \(AC - {B^2} \le 0\) (đpcm)