1. and
2. because
3. Although
4. so
5. that
6. why
7. and

The person that I gave it to was a very good friend of mine at the time. His name was Kaliya (1)___and_______ he came from India. I knew him (2)______because___ we studied together at a language school in Cambridge. (3)________Although____ we were from different backgrounds and cultures, we got on really well and we had the same sense of humor (4)______so_____ we became very good friends.

The present was a picture that had been painted of the River Cam in Cambridge. It was not so big - maybe around 10 inches by 14 inches, but it was very beautiful. It came in a gold plated frame and the picture had been drawn by a particularly well-known Cambridge artist. It had been signed by this person as well.

It’s the first time I have given this person a present, and I guess it’s quite different from presents (5)___which ______ I have given to other people before – I don’t recall ever giving someone a picture actually. If I’m buying for family then I’ll usually buy clothes or maybe some jewelry if it is a special occasion. Normally when I’ve bought something for friends it’s something more jokey so we can have a laugh about it, nothing that serious.

The reason (6)____why_____ I decided to give this particular gift is because we had spent a lot of time together in Cambridge and we had had some really fun times punting on the River Cam – that’s obviously why I thought this was an appropriate present. We used to go punting at least once a week, sometimes a couple of times. On one occasion there was a group of about ten of us that went down there, and we spent the whole day sitting by the river in the sun (7)_____and_____ as usual we went on a boat trip together. We all have a lot of photos to remind us of this great day.

So my friend, Kaaliya, is the person that I gave a present to and this was because I felt it would always remind him of the fun times that we had and also of Cambridge.

1. Đáp án B
2. Đáp án D
3. Đáp án D
4. Đáp án C
5. Đáp án B

1. He got wet_______ he forgot his umbrella.

A. because of    B. because

C. but     D. and

2. He stops working _______ heavy raining.

A. in spite of     B. although

C. despite     D. because of

3. They have a lot of difficulties in their life _______ their poverty.

A. in spite of     B. although

C. because     D. because of

4. Tom wakes his parents up_______ playing the guitar very softly.

A. because     B. in spite of

C. because of    D. although

5. Nobody could hear her_______ she spoke too quietly.

A. although     B. because

C. because of    D. in spite of

Ta có: \(A=\dfrac{2n+3}{8n}\) nguyên 

\(\Rightarrow4A=4\cdot\dfrac{2n+3}{8n}=\dfrac{8n+12}{8n}\) nguyên 

\(\Rightarrow4A=1+\dfrac{12}{8n}=1+\dfrac{3}{2n}\)  

Để 4A nguyên thì 3 ⋮ 2n 

⇒ 2n ∈ Ư(3) = {1; -1; 3; -3}

Mà: n ∈ Z ⇒ không có n thỏa mãn 

Tìm số nguyên n để thỏa A mãn điều gì vậy em?

\(A=92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)

\(=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)

\(=\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)

\(=8\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)

\(B=\dfrac{1}{45}+\dfrac{1}{50}+...+\dfrac{1}{500}\)

\(=\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)

=>\(\dfrac{A}{B}=\dfrac{8\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}=8:\dfrac{1}{5}=40\)

A = 1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7) + 1/(7.8)

= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8

= 1/3 - 1/8

= 5/24

1/(8.9) = 1/72

A = 5/24 ≠ 1/72

Em xem lại đề nhé

bạn xem lại biểu thức dưới mẫu nha

xin lỗi bạn nha mình bấm nhầm xóa số 6

a: Vì OA và OB là hai tia đối nhau

nên O nằm giữa A và B

b: O nằm giữa A và B

=>OA+OB=AB

=>AB=2+5=7(cm)

\(\dfrac{-3}{25}+\dfrac{9}{17}-\dfrac{22}{25}+\dfrac{25}{17}\)

\(=\left(-\dfrac{3}{25}-\dfrac{22}{25}\right)+\left(\dfrac{9}{17}+\dfrac{25}{17}\right)\)

\(=-\dfrac{25}{25}+\dfrac{34}{17}\)

=-1+2

=1

Lời giải:
Đặt $2n^2=ma$ với $a$ là số nguyên dương

$\Rightarrow m=\frac{2n^2}{a}$

$\Rightarrow n^2+m=n^2+\frac{2n^2}{a}$

Giả sử $n^2+m=n^2+\frac{2n^2}{a})$ là scp. Đặt $n^2+\frac{2n^2}{a}=k^2(k\in\mathbb{N})$
$\Rightarrow n^2a+2n^2=ak^2$

$\Rightarrow n^2(a+2)=ak^2$

$\Rightarrow n^2(a^2+2a)=a^2k^2=(ak)^2$

Mà $a^2+2a\in\mathbb{Z}^+$ nên $\Rightarrow a^2+2a$ cũng phải là 1 scp

Hiển nhiên $a^2+2a=(a+1)^2-1< (a+1)^2$ và $a^2+2a> a^2$

$\Rightarrow a^2< a^2+2a< (a+1)^2$

Theo định lý kẹp thì $a^2+2a$ không thể là scp. Tức là điều gs là vô lý.

$\Rightarrow n^2+m$ không là scp.

Chỉ nên tham khảo thôi:

Giả sử tồn tại n,m thỏa mãn \(n^2+m\) là số chính phương

Đặt \(m=\dfrac{2n^2}{p}\)

-> \(n^2+m=n^2+\dfrac{2n^2}{p}=n^2\left(1+\dfrac{2}{p}\right)\)

->\(1+\dfrac{2}{p}\) là bình phương một số hữu tỉ

->\(1+\dfrac{2}{p}=\dfrac{p+2}{p}=\dfrac{a^2}{b^2}\) với UCLN(a,b)=1 và a>b>0

->\(\left\{{}\begin{matrix}p+2=k\cdot a^2\\p=k\cdot b^2\end{matrix}\right.\)

->\(k\cdot\left(a^2-b^2\right)=2\)

Lại có p+2 và p chia hết cho k nên (p+2)-p=2 chia hết cho k

->k=1 hoặc k=2

TH1: k=1-> \(a^2-b^2=2\)

Nếu a,b cùng chẵn hoặc cùng lẻ thì \(a^2-b^2\) chia hết cho 4(vô lí)

Nếu a,b không cùng tính chẵn lẻ thì \(a^2-b^2\) lẻ (vô lí)

TH2: k=2-> \(a^2-b^2=1\)

-> a=1, b=0(vô lí)

Vậy giả sử sai, suy ra điều phải chứng minh

Lời giải:

Bạn Mai góp số phần tiền so với tổng số tiền là:
$\frac{1}{1+3}=\frac{1}{4}$

Bạn Hòa góp số phần tiền so với tổng số tiền là:

$\frac{1}{1+2}=\frac{1}{3}$
60000 đồng của Bình ứng với số phần tổng số tiền là:
$1-\frac{1}{4}-\frac{1}{3}=\frac{5}{12}$

Giá bộ sách tham khảo:

$60000:\frac{5}{12}=144000$ (đồng)

Số tiền Mai góp: $144000\times \frac{1}{4}=36000$ (đồng)

Số tiền Hòa góp: $144000\times \frac{1}{3}=48000$ (đồng)