Gọi 2 số chính phương liên tiếp đó là \(n^2,\left(n+1\right)^2\). Ta có:

\(P=n^2+\left(n+1\right)^2+n^2\left(n+1\right)^2\)

\(=n^2+n^2+2n+1+n^2\left(n^2+2n+1\right)\)

\(=n^4+2n^3+3n^2+2n+1\)

Ta có \(\dfrac{P}{n^2}=n^2+2n+3+\dfrac{2}{n}+\dfrac{1}{n^2}\)

\(=\left(n+\dfrac{1}{n}\right)^2+2\left(n+\dfrac{1}{n}\right)+1\)

\(=\left(n+\dfrac{1}{n}+1\right)^2\)

\(\Rightarrow P=\left[n\left(n+\dfrac{1}{n}+1\right)\right]^2=\left(n^2+n+1\right)^2=\left[n\left(n+1\right)+1\right]^2\)

 Dễ dàng kiểm chứng được \(2|n\left(n+1\right)\), do đó \(n\left(n+1\right)+1\) là số lẻ, suy ra đpcm.

Hai số chính phương liên tiếp là \(n^2;\left(n+1\right)^2\)

Theo đề ta có :

\(n^2+\left(n+1\right)^2+n^2\left(n+1\right)^2\)

\(=n^2+n^2+2n+1+n^4+2n^3+n^2\)

\(=\left(n^4+n^3+n^2\right)+\left(n^3+n^2+n\right)+\left(n^2+n+1\right)\)

\(=n^2\left(n^2+n+1\right)+n\left(n^2+n+1\right)+\left(n^2+n+1\right)\)

\(=n^2\left(n^2+n+1\right)+n\left(n^2+n+1\right)+\left(n^2+n+1\right)\)

\(=\left(n^2+n+1\right)^2\)

\(=\left[n\left(n+1\right)+1\right]^2\)

mà \(n\left(n+1\right)⋮2\) (là 2 số tự nhiên liên tiếp)

\(\Rightarrow n\left(n+1\right)+1\) là số lẻ

\(\Rightarrow\left[n\left(n+1\right)+1\right]^2\) là số chính phương lẻ

\(\Rightarrow dpcm\)

Ta có \(\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x+y\right)\left(x+4y\right)\left(x+2y\right)\left(x+3y\right)+y^4\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

\(=\left(x^2+5xy+5y^2-y^2\right)\left(x^2+5xy+5y^2+y^2\right)+y^4\)

\(=\left(x^2+5xy+5y^2\right)^2\) là số chính phương. \(\Rightarrowđpcm\)

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int n,k;
    cin >> n >> k;
    int a[n];
    for (int i=0;i<n;i++)
        cin >> a[i];
    sort(a,a+n);
    cout << a[k-1];
    return 0;
}

\(a_n=1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow a_{n+1}=1+2+3+...+n+\left(n+1\right)=\dfrac{\left(n+1\right)\left(n+2\right)}{2}\)

\(\Rightarrow a_n+a_{n+1}=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n+1\right)\left(n+2\right)}{2}\)

\(=\dfrac{\left(n+1\right)}{2}.\left(n+n+2\right)=\dfrac{\left(n+1\right)}{2}.\left(2n+2\right)\)

\(=\dfrac{\left(n+1\right)}{2}.2\left(n+1\right)=\left(n+1\right)^2\)

\(\Rightarrow dpcm\)

ko bt

 Pt đã cho \(\Leftrightarrow\left(\dfrac{10a+b}{a+b}\right)^2=a+b\inℤ\). Ta thấy nếu \(a+b\) không là số chính phương thì khi đó \(\sqrt{a+b}=\dfrac{10a+b}{a+b}\), vô lí vì VT là số vô tỉ trong khi VP là số hữu tỉ (do \(a,b\inℤ\)). Do đó, \(a+b\) phải là số chính phương hay \(\dfrac{10a+b}{a+b}=k\inℤ\) . Suy ra \(a+b=k^2\).

 Từ đó suy ra \(10a+b=k\left(a+b\right)=k^3\). Do đó ta có hệ pt sau:

 \(\left\{{}\begin{matrix}a+b=k^2\\10a+b=k^3\end{matrix}\right.\). Giải hpt, ta thu được họ nghiệm là \(\left(a,b\right)=\left(\dfrac{k^3-k^2}{9},\dfrac{10k^2-k^3}{9}\right)\). Do \(a,b\inℤ\) nên \(\left\{{}\begin{matrix}9|k^3-k^2\\9|10k^2-k^3\end{matrix}\right.\Leftrightarrow9|k^3-k^2\) \(\Leftrightarrow9|k^2\left(k-1\right)\). Hơn nữa \(\left(k^2,k-1\right)=1\) nên suy ra \(\left[{}\begin{matrix}9|k^2\\9|k-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3|k\\k\equiv1\left[9\right]\end{matrix}\right.\)

 Như vậy, tất cả các cặp số có dạng \(\left(\dfrac{k^3-k^2}{9},\dfrac{10k^2-k^3}{9}\right)\) với \(k⋮3\) hoặc \(k\equiv1\left[9\right]\) đều thỏa mãn pt đã cho.

 Ở dòng đầu tiên mình thiếu trường hợp nếu \(a+b=0\) thì \(10a+b=0\) \(\Leftrightarrow a=0\Leftrightarrow b=0\) là nghiệm của pt đã cho, sau đó mình xét \(a+b\ne0\) thì mới chia được 2 vế cho \(a+b\) như trong bài nhé.

Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)

hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)

\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)

\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))

\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)

\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)

\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)

\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)

\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow ac=bd\) (do \(b\ne d\))

 Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)

1, The bigger the apartment is, the higher the rent is

2, The sooner we set off, the sooner we will arrive

3, The better the joke is, the louder the laugher is

4, The fatter she gets, the more tired she feels

5, The older she gets, the less travel he wants

1. The bigger the apartment is, the higher the rent is.
2. The sooner we set off, the sooner we will arrive.
3. The better the joke is, the louder the laughter is.
4. The fatter she gets, the more tired she feels.
5. The older he gets, the less he wants to travel.
6. The more difficult the games are, the more excited the children are.
7. The faster people dive, the more accidents happen.
8. The more I meet him, the more I hate him.
9. The less time my boss is pressed for, the better he works.
10. The more money he has, the more he wants to spend.

1. The more books you read, the more knowledge you will have

2. The more he speaks, the more bored people feel

3. The more the economy grows, the better people's living condition is

4. The farther he traveled, the more he learned

1. C

2. B

3. D

4. A

5. D

6. D

7. C

8. B

9. C

1. C. better
2. B. as
3. D. tallness
4. A. as/as
5. D. A and B are correct
6. A. as/as
7. C. faster than
8. C. better
9. A. most
10. C. higher than

1. I've never tasted a more delicious cake.
2. George is the most dependable person I've ever met.
3. Your mother is the kindest-hearted person I know.
4. Mr John is the best teacher in this school.
5. Is this one bigger than that one?