Câu 7:
ĐK: $-1\le x\le 1$
$y=x.\sqrt{1-x^2}$
$y'=\sqrt{1-x^2}-\dfrac{x^2}{\sqrt{1-x^2}}$
$=\dfrac{1-2x^2}{\sqrt{1-x^2}}$
$y'=0\to x=\pm\dfrac{1}{\sqrt2}$
So sánh $f(-1); f(1); f\left(\dfrac{1}{\sqrt2}\right)$, $f\left(\dfrac{-1}{\sqrt2}\right)$
$\to \min\limits_{[-1;1]}f(x)=f\left( \dfrac{-1}{\sqrt2}\right)=\dfrac{-1}{2}$
$\to C$
Câu 10:
$y=x-1+\dfrac{1}{x-1}+3$
Theo BĐT AM-GM:
$y\ge 2.\sqrt1+3=5$
$\min\limits_{(1;+\infty)}y=5$
Dấu $=$ xảy ra khi $x=2$
$\to D$