MN ơi giúp mk vs ạ , mk sắp phải nộp r 😥
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Tờ 1
41 It's very important to use body language in communication
42 Despite her age, she still leads an active life
43 My mother said that you had to decorate the room carefully
44 People recycle old cans to make new ones
45 Tim is always forgetting his homework
46 T
47 F
48 T
49 T
50 F
Tờ 2
17 C => hard
18 do => make
19 D => has
20 to go => going
21 A => At
22 B => to
23 C => beautifully
24 D => five-star
25 is => was
V
26 would travel
27 be
28 to buy
29 has spoken
30 Has - just been finished
VI
31 for
32 as
33 about
34 with
35 than
VII
36 development
37 exploration
38 behavior
39 deforestation
40 specialness
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1) He always drives his car out of the garage at nine o'clock .
2) They find a parking place near the shops rarely
3) I sometimes fly with my parents to Florida in spring
4) She often comes to school late in the winter
5) They always meet their friends at the sports ground after dinner
6) She always enjoys swimming in our pool in the morning
7) On Mondays, my mother always does the washing
8) I put out the dustbins once a week .
9) We often go for a walk with our dog
10) My sister sometimes does the ironing ,
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9. Has the work been done by him?
10. The boxes were opened and cigarettes were taken out by us
11. She was given a new one
12. He is proved wrong
13. We were promised higher wages
14. This is the third time we have been written to about this by them
15. We were asked to be there at 8 o'clock
16. She is being shown how to do it
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a) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:
\(BC^2=AB^2+AC^2\)
\(\Leftrightarrow BC^2=3^2+4^2=25\)
hay BC=5(cm)
b) Xét ΔABC có AB<AC<BC(3cm<4cm<5cm)
mà góc đối diện với cạnh AB là \(\widehat{ACB}\)
và góc đối diện với cạnh AC là \(\widehat{ABC}\)
và góc đối diện với cạnh BC là \(\widehat{BAC}\)
nên \(\widehat{ACB}< \widehat{ABC}< \widehat{BAC}\)
Xét ΔABC có
HB là hình chiếu của AB trên BC
HC là hình chiếu của AC trên BC
AB<AC
Do đó: HB<HC
c) Xét ΔCAB vuông tại A và ΔCAD vuông tại A có
CA chung
AB=AD(gt)
Do đó: ΔCAB=ΔCAD(hai cạnh góc vuông)
Suy ra: CB=CD(hai cạnh tương ứng)
Xét ΔCBD có CB=CD(cmt)
nên ΔCBD cân tại C(Định nghĩa tam giác cân)
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c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
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1. that
2. that night
3. the day before.
4. my
5. that evening
6. the previous week
7. her
8. there
9. the following day
10. two hours
11. then
1.that
2.that night
3.the day before
4.my
5.that night
6.the previous week
7.her 8.there 9 the following day 10.two hours before 11.then
Xét ΔAMB có
MD là đường phân giác ứng với cạnh AB
nên \(\dfrac{AD}{DB}=\dfrac{AM}{MB}\)(1)
Xét ΔAMC có
ME là đường phân giác ứng với cạnh AC
nên \(\dfrac{AE}{EC}=\dfrac{AM}{MC}\)(2)
Ta có: M là trung điểm của BC(gt)
nên MB=MC(3)
Từ (1), (2) và (3) suy ra \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\)
hay DE//BC(đpcm)