EM THAM KHẢO:

$\textbf{Câu 16.1:}$

$y = \dfrac{-5x-4}{-x-1}$

$\Rightarrow y' = \dfrac{(-5)(-1) - (-4)(-1)}{(-x-1)^2}= \dfrac{5 - 4}{(-x-1)^2}= \dfrac{1}{(-x-1)^2}$

$\Rightarrow y'(-3) = \dfrac{1}{(-(-3)-1)^2}= \dfrac{1}{(3-1)^2} = \dfrac{1}{4}$

$\Rightarrow \text{Chọn C}$

$\textbf{Câu 16.2:}$

$y = \dfrac{5x+6}{7x+7}$

$\Rightarrow y' = \dfrac{5 \cdot 7 - 6 \cdot 7}{(7x+7)^2}= \dfrac{35 - 42}{(7x+7)^2}= \dfrac{-7}{(7x+7)^2}$

$\Rightarrow \text{Chọn C}$

$\textbf{Câu 16.3:}$

$y = \dfrac{2x+7}{x+4}$

$\Rightarrow y' = \dfrac{2 \cdot 4 - 7 \cdot 1}{(x+4)^2}= \dfrac{8 - 7}{(x+4)^2}= \dfrac{1}{(x+4)^2}$

$\Rightarrow y'(2) = \dfrac{1}{(2+4)^2}= \dfrac{1}{6^2}= \dfrac{1}{36}$

$\Rightarrow \text{Chọn A}$

$\textbf{Câu 16.4:}$

$y = \dfrac{5x-4}{x+1}$

$\Rightarrow y' = \dfrac{5 \cdot 1 - (-4) \cdot 1}{(x+1)^2}= \dfrac{5 + 4}{(x+1)^2}= \dfrac{9}{(x+1)^2}$

$\Rightarrow y'(-2) = \dfrac{9}{(-2+1)^2}= \dfrac{9}{(-1)^2}= 9$

$\Rightarrow \text{Chọn C}$

$\textbf{Câu 16.5:}$

$y = \dfrac{-2x-5}{-x-2}$

$\Rightarrow y' = \dfrac{(-2)(-2) - (-5)(-1)}{(-x-2)^2}= \dfrac{4 - 5}{(-x-2)^2}= \dfrac{-1}{(-x-2)^2}$

$\Rightarrow y'(-5) = \dfrac{-1}{(-(-5)-2)^2}= \dfrac{-1}{(5-2)^2}= \dfrac{-1}{3^2}= \dfrac{-1}{9}$

$\Rightarrow \text{Chọn C}$