$a) \ y= f(x)= -2x^{2}-7 $ trên  $(-4;0) $ và  $(3;10)$
$b) \ y=f(x)=  \frac{-1}{x-1} $trên $ (-3;-2) $ và $(2;3)$
a) Đạo hàm rồi xét dấu $y'$ thôi có gì mà hỏi chứ

TXĐ: $D=R$

$y'=-4x;\ y'= 0\Leftrightarrow x=0$ kẻ trục xét dấu ra ta thấy $y' >0 \forall x \in (-4;\ 0)$ nên hàm số đồng biến trên $(-4;\ 0)$

$y' <0 \forall x \in (3;\ 10)$ nên hàm số nghịc biến trên $(3;\ 10)$

Câu b tương tự

TXĐ: $D= R/ \{1\}$

 $y' =\dfrac{1}{(x-1)^2} > 0 \forall x \in D$

Do đó hàm số đồng biến trên 2 khoảng đã cho

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