Đáp án:
$A=2x^2+9x-13$
$=2x^2+2.\sqrt{2}x.\sqrt{2}.\dfrac{9}{4}+\dfrac{81}{8}-\dfrac{185}{8}$
$=\left(x\sqrt{2}+\dfrac{9}{4}.\sqrt{2}.\right)^2-\dfrac{185}{8}$
Do $\left(x\sqrt{2}+\dfrac{9}{4}.\sqrt{2}.\right)^2\geq 0$
$\Rightarrow A\geq -\dfrac{185}{8}$
Dấu "=" xảy ra khi và chỉ khi $x\sqrt{2}+\dfrac{9}{4}.\sqrt{2}=0$
$\Rightarrow x=-\dfrac{9}{4}$
Vậy $A_{min}=-\dfrac{185}{8}$ khi $x=-\dfrac{9}{4}$
$B=-\dfrac{-x^2+x-10}{x^2-2x+1}$
$B=-\dfrac{-x^2+2x-1-x+1-10}{(x-1)^2}$
$B=-\dfrac{-(x-1)^2-(x-1)-10}{(x-1)^2}$
$B=1+\dfrac{1}{x-1}+\dfrac{10}{(x-1)^2}$
Đặt $\dfrac{1}{x-1}=a$, ta có
$B=1+a+10a^2$
$B=10a^2+a+\dfrac{1}{40}+\dfrac{39}{40}$
$B=\left(a\sqrt{10}+\dfrac{1}{\sqrt{40}}\right)^2+\dfrac{39}{40}$
$B\geq \dfrac{39}{40}$
Dấu "=" xảy ra khi $a\sqrt{10}+\dfrac{1}{\sqrt{40}}=0$
$\Rightarrow a=-\dfrac{1}{40\sqrt{10}}$
$\Rightarrow \dfrac{1}{x-1}=-\dfrac{1}{40\sqrt{10}}$
$\Rightarrow x=1-40\sqrt{10}$
Vậy $B_{min}=\dfrac{39}{40}$ khi $x=1-40\sqrt{10}$
