tìm số dư trong phép chia 2006^2024 cho 7
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S
subjects
CTVHS
9 tháng 3 2024
X. Complete the second sentence
38 Despite the good weather, we stayed indoors
39 When it began to rain, the children were playing in the playground
40 Before she broke the eggs, she mixed the yolk with flour
41 We won't be able to make coffee until the coffee machine is repaired
42 While I am watching TV, my brother is playing video games
XI. Use suggested words to write sentences
43 Last week, when I went down the street, a small cat suddenly jumped out on the street
44 She thinks I need to focus more on my schoolwork than on the pet
45 My sister once rescued a dog when it was badly injured by its owner
46 We have plans to build an animal hospital in the future
SV
Sinh Viên NEU
CTVVIP
29 tháng 2 2024
16. There are only 50 saolas around Viet Nam and they are the most critically _________ animals in Viet Nam.
A. threatening B. popular C. endangered D. dangerous
17. The sun _________ at 5 a.m. tomorrow according to the weather forecast.
A. is going to rise B. rises C. rise D. will rise
18. Local officials posted a flood __________ yesterday, so many are moving to safer areas and using more flood-prevention methods today. A. damage B. warning C. property D. emergency
19. While my mother __________ dinner, my father _________ home from work yesterday.
A. was made/ came B. was making/ came C. was making/ was coming D. made/ came
20. I __________ on the sofa when suddenly the cat ___________ on me.
A. was sitting/ jumped B. was sitting/ was jumping C. sat/ jumped D. sat/ was jumping
21. My dentist __________ checks my teeth. I don’t often go to the dentist.
A. never B. sometimes C. always D. usually
22. The students ___________ on their field trip to the Happy Farm next day.
A. will go B. goes C. go D. is going
23. I have to pass the entrance test ________ I can apply to that university.
A. after B. as soon as C. while D. before
24. My family was having dinner _________ it was raining outside.
A. till B. while C. until D. as soon as
29 tháng 2 2024
16c
17a
18b
19b
20a
21b
22d
23d
24b
Đáp án nhé e chúc e may mắn chạy deadline
Trả công a nha
NL
Nguyễn Lê Phước Thịnh
CTVHS
29 tháng 2 2024
a: Xét ΔNHE vuông tại E và ΔNMH vuông tại H có
\(\widehat{HNE}\) chung
Do đó: ΔNHE~ΔNMH
=>\(\dfrac{NH}{NM}=\dfrac{NE}{NH}\)
=>\(NH^2=NE\cdot NM\left(1\right)\)
Xét ΔHFN vuông tại F và ΔPHN vuông tại H có
\(\widehat{HNF}\) chung
Do đó: ΔHFN~ΔPHN
=>\(\dfrac{NH}{NP}=\dfrac{NF}{NH}\)
=>\(NH^2=NP\cdot NF\left(2\right)\)
Từ (1),(2) suy ra \(NE\cdot NM=NP\cdot NF\)
b: Ta có: \(NE\cdot NM=NP\cdot NF\)
=>\(\dfrac{NE}{NP}=\dfrac{NF}{NM}\)
Xét ΔNEF và ΔNPM có
\(\dfrac{NE}{NP}=\dfrac{NF}{NM}\)
\(\widehat{ENF}\) chung
Do đó: ΔNEF~ΔNPM
=>\(\widehat{NEF}=\widehat{NPM}\)
c: ta có: \(\widehat{NEF}=\widehat{NPM}\)
mà \(\widehat{NEF}=\widehat{KEM}\)(hai góc đối đỉnh)
nên \(\widehat{KEM}=\widehat{KPN}\)
Xét ΔKEM và ΔKPF có
\(\widehat{KEM}=\widehat{KPF}\)
\(\widehat{EKM}\) chung
Do đó: ΔKEM~ΔKPF
=>\(\dfrac{KE}{KP}=\dfrac{KM}{KF}\)
=>\(KE\cdot KF=KM\cdot KP\)
HM
1
P
29 tháng 2 2024
Pt: \(\dfrac{3}{x^2+x+1}+\dfrac{4}{x^2+x+2}-\dfrac{6}{x^2+x+4}=1\) (*)
ĐK: \(\left\{{}\begin{matrix}x^2+x+1\ne0\\x^2+x+2\ne0\\x^2+x+4\ne0\end{matrix}\right.\)(luôn đúng)
Đặt: \(x^2+x+2=t\ge\dfrac{7}{4}\)
(*) trở thành:
\(\dfrac{3}{t-1}+\dfrac{4}{t}-\dfrac{6}{t+2}=1\)
\(\Leftrightarrow\dfrac{3t\left(t+2\right)}{t\left(t-1\right)\left(t+2\right)}+\dfrac{4\left(t-1\right)\left(t+2\right)}{t\left(t-1\right)\left(t+2\right)}-\dfrac{6t\left(t-1\right)}{t\left(t-1\right)\left(t+2\right)}=1\)
\(\Leftrightarrow3t\left(t+2\right)+4\left(t-1\right)\left(t+2\right)-6t\left(t-1\right)=t\left(t-1\right)\left(t+2\right)\)
\(\Leftrightarrow3t^2+6t+4\left(t^2+t-2\right)-6t^2+6t=t\left(t^2+t-2\right)\)
\(\Leftrightarrow-3t^2+12t+4t^2+4t-8=t^3+t^2-2t\)
\(\Leftrightarrow t^2+16t-8=t^3+t^2-2t\)
\(\Leftrightarrow t^3-18t+8=0\)
\(\Leftrightarrow\left(t-4\right)\left(t^2+4t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=4\left(tm\right)\\t=\sqrt{6}-2\left(ktm\right)\\t=-\sqrt{6}-2\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: ...
NV
8
HM
0
concặc
Ta có \(2006^{2024}=\left(7.286+4\right)^{2024}\) \(=7A+4^{2024}\). Do đó ta chỉ cần tìm số dư của \(4^{2024}\) khi chia cho 7.
Để ý rằng: \(4^0\equiv1\left[7\right]\); \(4^1\equiv4\left[7\right]\); \(4^2\equiv2\left[7\right]\); \(4^3\equiv1\left[7\right]\); \(4^4\equiv4\left[7\right]\); \(4^5\equiv2\left[7\right]\)
Do đó ta nảy sinh dự đoán rằng \(4^{3k+2}\equiv2\left[7\right]\left(k\inℕ\right)\). Ta sẽ chứng minh điều này bằng phương pháp quy nạp,
Thật vậy, với \(k=0\) thì khẳng định đúng (theo như trên)
Giả sử khẳng định đúng đến \(k=l\ge0\), khi đó \(4^{3l+2}\equiv2\left[7\right]\). Ta cần chứng minh khẳng định đúng với \(k=l+1\), tức là cm \(4^{3\left(l+1\right)+2}\equiv2\left[7\right]\)
Thật vậy, ta có \(4^{3\left(l+1\right)+2}\equiv4^{3l+3+2}\equiv64.4^{3l+2}\equiv1.2\equiv2\left[7\right]\)
Vậy khẳng định đúng với \(k=l+1\Rightarrow4^{3k+2}\equiv2\left[7\right]\)
Vì vậy \(4^{2024}=4^{2022+2}=4^{3.674+2}\equiv2\left[7\right]\)
Vậy số dư của phép chia \(2006^{2024}\) cho 7 là 2.