1D 2D 3D 4A 5A 6D 7A 8B 9D 10A 11B 12C 13A 14D 15C 16B 17C 18B

Giúp em bài số 4 ạ😓

1 Having slept

2 not being invited

3 Having had

4 having

5 talking

6 succeeded - launching

7 Having travelled

8 Have - considered - trying

9 Having seen - had - to go

10 Being invited

11 Being found

12 having

13 taken - being photographed 

14 to fix

15 living

16 Having waited - to deliver - decided to cancel

17 Having photocopied 

18 to have happen

19 to give

20 spoiling

1 Jill reminded John to do the washing up

2 The police ordered his men to search all the shops on that street

3 She blamed me for ignoring the notice about life-saving equipment

4 My aunt advised me not to argut with my father

5 Stella congratulated Jeff on having got an promotion at last

6 Kevin apoligized to Sarah for making her angry

7 The man warn his son to put down the gun

8 Ron denied being in the town on the night of the robbery

9 Ted promise to pay back the money at the end of that month

10 George encouraged Susan to send her story to the magazine

11 Natalie accused Tom of lying to her

D C D B C C C D A 

Sam puts up the decorations.

Five guests came to the party.

Molly and Sam play video games with their cousins

The family was celebrating too early. Dad's birthday was two months away.

relationship

celebration

married

golden

quietly

celebratory

refer

1. he dislike being called " the liar " => He dislike people..CALLING HIM THE LIAR.

2. The police are following the suspects => The suspects ..ARE BEING FOLLOWED BY THE POLICE.

3. She always expects to be admired by everybody => She always expects everybody..TO ADMIRE HER...

4. Someone stole his car two days ago => He had ..HIS CAR STOLEN BY SOMEONE TWO DAYS AGO..

2:

a: pi/2<a<pi

=>cosa<0

sin^2a+cos^2a=1

=>cos^2a=1-4/9=5/9

=>cosa=-căn 5/3

cos2a=2*cos^2a-1=2*5/9-1=10/9-1=1/9

sin(2a-pi/6)

=sin2a*cospi/6-cos2a*sinpi/6

=2*sina*cosa*(căn 3/2)-1/9*1/2

\(=2\cdot\dfrac{2}{3}\cdot\dfrac{-\sqrt{5}}{3}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{18}=\dfrac{-4\sqrt{15}-1}{18}\)

b; tan a=2

=>sin a=2*cosa

\(A=\dfrac{3\cdot\left(2\cdot cosa\right)^2-cos^2a+2}{5\cdot\left(2\cdot cosa\right)^2+3cosa\cdot2cosa}\)

\(=\dfrac{12\cdot cos^2a-cos^2a+2}{20cos^2a+6cos^2a}\)

\(=\dfrac{11cos^2a+2\left(4cos^2a+cos^2a\right)}{26cos^2a}=\dfrac{21}{26}\)

4:

a: (C): x^2+y^2-4x+2y-4=0

=>x^2-4x+4+y^2+2y+1=9

=>(x-2)^2+(y+1)^2=9

=>I(2;-1); R=3

b: Gọi (d) là phương trình cần tìm

(d)//4x+3y-1=0

=>(d): 4x+3y+c=0

I(2;-1);R=3

Theo đề, ta có: d(I;(d))=R=3

=>\(\dfrac{\left|4\cdot2+3\cdot\left(-1\right)+c\right|}{\sqrt{4^2+3^2}}=3\)

=>|c+5|=15

=>c=10 hoặc c=-20

a.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{6}}{3}\)

\(cos\left(x+\dfrac{\pi}{3}\right)=cosx.cos\left(\dfrac{\pi}{3}\right)-sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{6}-3}{6}\)

b.

\(\pi< x< \dfrac{3\pi}{2}\Rightarrow sinx< 0\)

\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\dfrac{5}{13}\)

\(B=sin\left(\dfrac{\pi}{3}-x\right)=sin\left(\dfrac{\pi}{3}\right).cosx-cos\left(\dfrac{\pi}{3}\right).sinx=...\) (bạn tự thay số bấm máy)

c.

\(A=cos^2x+cos^2y+2cosx.cosy+sin^2x+sin^2y+2sinx.siny\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)+2\left(cosx.cosy+sinx.siny\right)\)

\(=1+1+2cos\left(x-y\right)\)

\(=2+2cos\left(\dfrac{\pi}{3}\right)=...\)

d.

\(B=cos^2x+sin^2y+2cosx.siny+cos^2y+sin^2x-2sinx.cosy\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)-2\left(sinx.cosy-cosx.siny\right)\)

\(=2-2sin\left(x-y\right)=2-2sin\left(\dfrac{\pi}{3}\right)=...\)

Anh nghĩ với bài kiểm tra em nên tự làm nhé. 

f, \(3sin^2x-cosx+2cos2x-3=0\)

\(\Leftrightarrow3-3cos^2x-cosx+2\left(2cos^2x-1\right)-3=0\)

\(\Leftrightarrow cos^2x-cosx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pi+k2\pi\)

h, \(cos^2x+cos^22x+cos^23x+cos^24x=2\)

\(\Leftrightarrow2cos^2x+2cos^22x+2cos^23x+2cos^24x=4\)

\(\Leftrightarrow cos2x+cos4x+cos6x+cos8x=0\)

\(\Leftrightarrow2cos5x.cos3x+2cos5x.cosx=0\)

\(\Leftrightarrow cos5x\left(cos3x+cosx\right)=0\)

\(\Leftrightarrow2cos5x.cos2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cos2x=0\\cosx=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k\pi}{5}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

g, \(cos^4x-sin^4x=2cosx-1\)

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=2cosx-1\)

\(\Leftrightarrow cos2x-2cosx+1=0\)

\(\Leftrightarrow2cos^2x-2cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)

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

 1 had better eat more fruits and vegetables.

2 likes painting very much.

3 play  card.

*Le's -> Let's

4 had better not eat canned  food.

5 don't we go camping for some days?