Toán 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 05/10/2021 By Gabriella 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) Trả lời
Mình đang cần điểm ….thông cảm cho mình nha
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)