6 How often does your grandfather often listen to the radio

7 How often do you go to the gym

8 What did they send

9 What time did he leave the party

10 Where did he go

11 Who did he meet for coffee

12 Why did she called the firemen

13 How did he go to school

14 What does lan like doing

15 Where was your father last months

16 How much is the dictionary

17 How long did it take to finish the composition

18 Who bought a poster

What did Mrs Robinson by

19 Where is a restaurant

20 What did her neighbor give her

21 How was the homework yesterday

22 Do you have magaazine and newspapers

23 How often do you go to the movies

31)B

32)B

33)C

34)A

35)B

Câu 2

a) nH₂ = \(\dfrac{4,48}{22,4}=0,2mol\)

n_ancol = 2n_H2 = 0,4 mol

=> M_ancol = \(\dfrac{17}{0,4}=42,5\)

=> Hai ancol là CH₃OH và C₂H₅OH

b) số ete thu được tối đa = \(\dfrac{n\left(n+1\right)}{2}\)

= \(\dfrac{2.3}{2}=3\)

5. 

ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)

Pt tương đương:

\(\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow x=30^0+k180^0\)

6.

\(\Leftrightarrow2\sqrt{2}sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

5. \(\Leftrightarrow\left[{}\begin{matrix}\tan\left(x-30\right)=0\\\cos\left(2x-150\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-30=360k\\\left[{}\begin{matrix}2x-150=90+360k\\2x-150=270+360k\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30+360k\\x=120+180k\\x=210+180k\end{matrix}\right.\)

Vậy ...

6, \(\Leftrightarrow2\sqrt{2}\sin x.\cos x+2\cos x=0\)

\(\Leftrightarrow\cos x\left(1+\sqrt{2}\sin x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\sqrt{2}\sin x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{3}{2}\pi+k2\pi\end{matrix}\right.\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{7}{4}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

6.

\(\Leftrightarrow x^2+4x+3>m\) ; \(\forall x>1\)

\(\Leftrightarrow m< \min\limits_{x>1}\left(x^2+4x+3\right)\)

Xét hàm \(f\left(x\right)=x^2+4x+3\) với \(x>1\)

\(-\dfrac{b}{2a}=-2< 1\) ; \(f\left(1\right)=8\Rightarrow f\left(x\right)>8\) ; \(\forall x>1\)

\(\Rightarrow m\le8\)

7.

Do C thuộc d nên tọa độ có dạng: \(C\left(-2c-1;c\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;6\right)\\\overrightarrow{CA}=\left(2c;1-c\right)\end{matrix}\right.\)

\(AB\perp AC\Leftrightarrow\overrightarrow{AB}.\overrightarrow{AC}=0\Leftrightarrow4.2c+4\left(1-c\right)=0\)

\(\Leftrightarrow4c+4=0\Rightarrow c=-1\Rightarrow C\left(1;-1\right)\)

b.

 \(AB=\sqrt{4^2+6^2}=2\sqrt{13}\)

Phương trình đường thẳng AB qua A và nhận \(\left(3;-2\right)\) là 1 vtpt có dạng:

\(3\left(x+1\right)-2\left(y-1\right)=0\Leftrightarrow3x-2y+5=0\)

Do d thuộc d nên tọa độ có dạng: \(D\left(-2d-1;d\right)\)

\(S_{ABD}=\dfrac{1}{2}AB.d\left(D;AB\right)=50\)

\(\Leftrightarrow\dfrac{\sqrt{13}\left|3\left(-2d-1\right)-2d+5\right|}{\sqrt{3^2+\left(-2\right)^2}}=50\)

\(\Leftrightarrow\left|-8d+2\right|=50\Rightarrow\left[{}\begin{matrix}d=-6\\d=\dfrac{13}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}D\left(11;-6\right)\\D\left(-14;\dfrac{13}{2}\right)\end{matrix}\right.\)

2.

a, Gọi \(C=\left(-2m-1;m\right)\) là điểm cần tìm

\(AB=2\sqrt{13};AC=\sqrt{5m^2-2m+1};BC=\sqrt{5m^2+2m+65}\)

Ta có \(BC^2=AB^2+AC^2\)

\(\Leftrightarrow5m^2+2m+65=52+5m^2-2m+1\)

\(\Leftrightarrow m=-3\)

\(\Rightarrow C=\left(5;-3\right)\)

b, Gọi \(D=\left(-2n-1;n\right)\) là điểm cần tìm

Đường thẳng AB có phương trình \(\dfrac{x+1}{4}=\dfrac{y-1}{6}\Leftrightarrow3x-2y+5=0\)

Khoảng cách từ \(D\) đến \(AB\):

\(d\left(D;AB\right)=\dfrac{\left|3\left(-2n-1\right)-2n+5\right|}{\sqrt{3^2+2^2}}=\dfrac{\left|-8n+2\right|}{\sqrt{13}}\)

\(S_{ABC}=\dfrac{1}{2}.\dfrac{\left|-8n+2\right|}{\sqrt{13}}.2\sqrt{13}=50\)

\(\Rightarrow\left|4n-1\right|=25\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-6\\n=\dfrac{13}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}N=\left(11;-6\right)\\N=\left(-14;\dfrac{13}{2}\right)\end{matrix}\right.\)

ĐỔi 30 phút =0,5 h

a,Vận tốc của người đó 

\(v=\dfrac{s}{t}=\dfrac{5}{0,5}=10\left(\dfrac{km}{h}\right)\)

b,Chọn gốc tọa độ O trùng điểm cách A 1km và nằm trong AB

Chiều dương trục Ox : từ A đến B

Phương trình chuyển động của mỗi vật:

\(x_1=1+10t(km,h)\)

5:

a: góc ABO+góc ACO=180 độ

=>ABOC nội tiếp

b: Xét ΔABE và ΔADB có

góc ABE=góc ADB

góc BAE chung

=>ΔABE đồng dạng với ΔADB

=>AB^2=AE*AD

Xet (O) có

AB,AC là tiếp tuyến

=>AB=AC

mà OB=OC

nên OA là trung trực của BC

=>AH*AO=AB^2=AE*AD

=>AE/AO=AH/AD

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH:

Zn + 2HCl ---> ZnCl₂ + H₂

0,1<-0,2<----------------0,1

\(n_{ZnO}=\dfrac{10,55-0,1.65}{81}=0,05\left(mol\right)\)

ZnO + 2HCl ---> ZnCl₂ + H₂

0,05-->0,1

\(C_{M\left(HCl\right)}=\dfrac{0,1+0,2}{0,2}=1.5M\)