ta có :

y^' = -3x² -4x +1 → y^" = -6x -4 = 0

→ x = \(\dfrac{-2}{3}\)

A

b) phương trình
C₂H₅OH + CuO \(\underrightarrow{t^o}\) CH₃COOH + Cu + H₂O

C₃H₇OH + CuO \(\underrightarrow{t^o}\) C₂H₅COOH + Cu + H₂O

a) n_H2 = 0,125 mol

C₂H₅OH + Na → C₂H₅ONa + \(\dfrac{1}{2}\)H₂
x.............................................\(\dfrac{x}{2}\)

C₃H₇OH + Na → C₃H₇ONa + \(\dfrac{1}{2}\)H₂

y..............................................\(\dfrac{y}{2}\)

ta có \(\left\{{}\begin{matrix}46x+60y=12,2\\x+y=0,25\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0,2\\y=0,05\end{matrix}\right.\)

→%_C2H5OH = \(\dfrac{0,2.46}{12,2}.100\%\) \(\approx\) 75,41%

→%_C3H7OH = 24,59%

g^'(x) = \(\sqrt{x+3}\) 

ta có phương trình : \(\sqrt{x+3}\)  + \(\sqrt{2x-1}\) =3 ( ĐK : x\(\ge\)\(\dfrac{1}{2}\))

\(\Leftrightarrow\) x+3 +2x-1 +\(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 9

\(\Leftrightarrow\) \(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 7-3x

\(\Leftrightarrow\) 4(2x² +5x -3) = 49 - 42x +9x² 

\(\Leftrightarrow\) x² - 62x +61 = 0 \(\left\{{}\begin{matrix}x=61\\x=1\end{matrix}\right.\)

x_o = 2 → y_o = 2 

y^' = \(\dfrac{x-1}{\sqrt{x^2-2x+4}}\) → y^'(2) = \(\dfrac{1}{2}\)

phương trình tiếp tuyến có dạng: 

y = \(\dfrac{1}{2}\)x +1

f^'(x) = 3x² - 6x

-> f^'(3) = 3.3³ - 6.3 = 9 

đáp án D

y^' = \(\dfrac{\left(2x+2\right)\left(x-2\right)-\left(x^2+2x\right)}{\left(x-2\right)^2}\) = \(\dfrac{x^2-4x-4}{\left(x-2\right)^2}\)

→ y^'(1) = -7

a)ta có :
D = \(\dfrac{1}{f}\) = 3 (dp) → f =\(\dfrac{1}{3}m\)= \(\dfrac{100}{3}cm\)

b)d₁ = 10cm

→d₁^' = \(\dfrac{d_1.f}{d_1-f}\) = \(\dfrac{10.\dfrac{100}{3}}{10-\dfrac{100}{3}}\)= \(\dfrac{-100}{7}\) cm 

→ảnh là ảnh ảo

K = \(\dfrac{d_1'}{d_1}\) = \(\dfrac{-100}{7.10}\) = \(\dfrac{-10}{7}\)

b) vì pttt song song với đường thẳng y=-\(\dfrac{1}{8}\)x +5

→ K =- ​​​\(\dfrac{1}{8}\) ​

y^' = \(\dfrac{-2}{\left(x-1\right)^2}\) = \(\dfrac{-1}{8}\) → \(x\left[{}\begin{matrix}xo=5\\xo=-3\end{matrix}\right.\) → \(\left[{}\begin{matrix}y_o=\dfrac{3}{2}\\y_o=\dfrac{1}{2}\end{matrix}\right.\) 

- phương trình tiếp tuyến có dạng :

+ y = \(\dfrac{-1}{8}\)(x-5) +\(\dfrac{3}{2}\) = \(\dfrac{-1}{8}x\) + \(\dfrac{17}{8}\)

+ y= \(\dfrac{-1}{8}\)(x+3) +\(\dfrac{1}{2}\) = \(\dfrac{-1}{8}x\) + \(\dfrac{1}{8}\)