1.

 \(C_{12}H_{22}O_{11}+H_2O\rightarrow C_6H_{12}O_6+C_6H_{12}O_6\)

\(C_6H_{12}O_6\xrightarrow[xt:enzim]{t^o:30-35^o}2C_2H_5OH+2CO_2\)

\(C_2H_5OH+O_2\underrightarrow{xt:men\left(giấm\right)}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_4đặc,t^o}CH_3COOC_2H_5+H_2O\)

1.

 C12H22O11+H2O→C6H12O6+C6H12O6C12H22O11+H2O→C6H12O6+C6H12O6

C6H12O6to:30−35o−−−−−→xt:enzim2C2H5OH+2CO2C6H12O6→xt:enzimto:30−35o2C2H5OH+2CO2

C2H5OH+O2xt:men(giấm)−−−−−−−−−−−−→CH3COOH+H2OC2H5OH+O2xt:men(giấm)→CH3COOH+H2O

CH3COOH+C2H5OHH2SO4đặc,to−−−−−−−−−−→CH3COOC2H5+H2O

9C

10B

11D

12A

13B

14C

15B

16B

1.C

2.D

3.C

4.C

5.C

6.C

7.D

8.D

9.C

10.D

11.A

12.A

1.

\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)

\(\Rightarrow sin\alpha=-\sqrt{1-cos^2a}=-\dfrac{12}{13}\)

\(\Rightarrow tan2a=\dfrac{sin2a}{cos2a}=\dfrac{2sina.cosa}{cos^2a-sin^2a}=\dfrac{2.\left(-\dfrac{12}{13}\right).\left(\dfrac{5}{13}\right)}{\left(\dfrac{5}{13}\right)^2-\left(-\dfrac{12}{13}\right)^2}=...\)

3.

\(P=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x}+\dfrac{4}{4y}\ge\dfrac{\left(1+2\right)^2}{x+4y}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(\left(x;y\right)=\left(2;1\right)\)

4.

Lưu ý: hàm \(sinx\) đồng biến khi \(0< x< 90^0\) và nghịch biến khi \(90^0< x< 180^0\), hàm cos nghịch biến khi \(0< x< 90^0\)

Đường tròn (C) tâm \(I\left(1;1\right)\) bán kính \(R=4\) , \(\overrightarrow{IA}=\left(1;-1\right)\Rightarrow IA=\sqrt{2}\)

Theo công thức diện tích tam giác:

\(S_{IMN}=\dfrac{1}{2}IM.IN.sin\widehat{MIN}=\dfrac{1}{2}R^2.sin\widehat{MIN}=8.sin\widehat{MIN}\)

\(\Rightarrow S_{IMN}\) đạt max khi \(sin\widehat{MIN}\) đạt max

Gọi H là trung điểm MN \(\Rightarrow IH\perp MN\Rightarrow IH\le IA\) theo định lý đường xiên - đường vuông góc

\(\Rightarrow cos\widehat{HIM}=\dfrac{IH}{IM}\le\dfrac{IA}{IM}=\dfrac{\sqrt{2}}{4}\Rightarrow\widehat{HIM}>69^0\)

\(\Rightarrow\widehat{MIN}=2\widehat{HIM}>120^0>90^0\)

\(\Rightarrow sin\widehat{MIN}\) đạt max khi \(\widehat{MIN}\) đạt min

\(\Rightarrow\widehat{HIM}=\dfrac{1}{2}\widehat{MIN}\) đạt min

\(\Rightarrow cos\widehat{HIM}\) đạt max

\(\Rightarrow cos\widehat{HIM}=\dfrac{\sqrt{2}}{4}\Leftrightarrow H\) trùng A

Hay đường thẳng MN vuông góc IA \(\Rightarrow\) MN nhận (1;-1) là 1 vtpt

Phương trình MN: \(1\left(x-2\right)-1\left(y-0\right)=0\Leftrightarrow x-y-2=0\)

1.

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(Fe\left(OH\right)_2\rightarrow FeO+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)