Đáp án:
C39: $C\left( {\dfrac{7}{6};\dfrac{{13}}{6}} \right)$
Giải thích các bước giải:
C38:
ĐK: $a,b,c>0$
Ta có:
$\begin{array}{l}
\left( {a + b + c} \right)\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
= \left( {1 + \dfrac{a}{b} + \dfrac{a}{c}} \right) + \left( {\dfrac{b}{a} + 1 + \dfrac{b}{c}} \right) + \left( {\dfrac{c}{a} + \dfrac{c}{b} + 1} \right)\\
= 3 + \left( {\dfrac{a}{b} + \dfrac{b}{a}} \right) + \left( {\dfrac{a}{c} + \dfrac{c}{a}} \right) + \left( {\dfrac{b}{c} + \dfrac{c}{b}} \right)\\
\ge 3 + 2\sqrt {\dfrac{a}{b}.\dfrac{b}{a}} + 2\sqrt {\dfrac{a}{c}.\dfrac{c}{a}} + 2\sqrt {\dfrac{b}{c}.\dfrac{c}{b}} \left( {BDT:Cauchy} \right)\\
= 3 + 2 + 2 + 2\\
= 9
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{a}{b} = \dfrac{b}{a}\\
\dfrac{a}{c} = \dfrac{c}{a}\\
\dfrac{b}{c} = \dfrac{c}{b}
\end{array} \right. \Leftrightarrow a = b = c$
Vậy ta có điều phải chứng minh.
C39:
Ta có:
$C\in \Delta $ nên $C\left( {1 + t;2 + t} \right)$
Để $\Delta ACB$ cân ở $C$
$\begin{array}{l}
\Leftrightarrow CA = CB\\
\Leftrightarrow C{A^2} = C{B^2}\\
\Leftrightarrow {\left( {1 + t + 1} \right)^2} + {\left( {2 + t - 2} \right)^2} = {\left( {1 + t - 3} \right)^2} + {\left( {2 + t - 1} \right)^2}\\
\Leftrightarrow {\left( {t + 2} \right)^2} + {t^2} = {\left( {t - 2} \right)^2} + {\left( {t + 1} \right)^2}\\
\Leftrightarrow 2{t^2} + 4t + 4 = 2{t^2} - 2t + 5\\
\Leftrightarrow 6t = 1\\
\Leftrightarrow t = \dfrac{1}{6}\\
\Rightarrow C\left( {\dfrac{7}{6};\dfrac{{13}}{6}} \right)
\end{array}$
Vậy $C\left( {\dfrac{7}{6};\dfrac{{13}}{6}} \right)$
