Đáp án:
1)
Ta có:
$\dfrac{2a^2-10a-1}{a^2-2a+1}=\dfrac{(2a^2-4a+2)-(6a-6)-9}{(a-1)^2}$
$=\dfrac{2(a-1)^2-6(a-1)-9}{(a-1)^2}=2-\dfrac{6}{a-1}-\dfrac{9}{(a-1)^2}$
$=3-\left(1+\dfrac{2.3}{a-1}+\left(\dfrac{3}{a-1}\right)\right)$
$=3-\left(1+\dfrac{3}{a-1}\right)^2$
Do $\left(1+\dfrac{3}{a-1}\right)^2 \geq 0 \forall a\neq 1$ nên $3-\left(1+\dfrac{3}{a-1}\right)^2 \leq 3$.
Dấu "=" xảy ra khi và chỉ khi $1+\dfrac{3}{a-1}=0 \Leftrightarrow \dfrac{a-1+3}{a-1}=0 \Leftrightarrow a-2 =0 \Leftrightarrow a=2 (tm)$
Vậy $D_{max}=3 \Leftrightarrow a=2 $.
2)
Ta có:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}$
$\Leftrightarrow a^2c+b^2a+c^2b=b^2c+c^2a+a^2b$
$\Leftrightarrow(a^2c-a^2b)+(b^2a-c^2a)+(c^2b-b^2c)=0$
$\Leftrightarrow a^2(c-b)-a(c-b)(c+b)+bc(c-b)=0$
$\Leftrightarrow(c-b)(a^2-ac-ab+bc)=0$
$\Leftrightarrow\left[ \begin{array}{l}c-b=0\\a^2-ac-ab+bc=0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}c=b\\a^2-ac-ab+bc=0\end{array} \right.$
$\Rightarrow đpcm$
