\(Q=\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)=\)

\(=a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1=\)

\(=a^2b^2c^2+\left(a^2b^2+b^2c^2+a^2c^2\right)+\left(a^2+b^2+c^2\right)+1\) (1)

Ta có

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\)

\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1-2abc\left(a+b+c\right)\) (2)

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=\)

\(=a^2+b^2+c^2+2\)

\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\) (3)

Thay (2) và (3) vào (1)

\(Q=a^2b^2c^2+1-2abc\left(a+b+c\right)+\left(a+b+c\right)^2-2+1=\)

\(=\left(abc\right)^2-2abc\left(a+b+c\right)+\left(a+b+c\right)^2=\)

\(=\left[abc-\left(a+b+c\right)\right]^2\)

\(n=a^2+b^2\)

\(\Rightarrow n^2=\left(a^2+b^2\right)^2-4a^2b^2+4a^2b^2=\)

\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)+\left(2ab\right)^2=\)

\(=\left(a-b\right)^2\left(a+b\right)^2+\left(2ab\right)^2=\)

\(=\left[\left(a-b\right)\left(a+b\right)\right]^2+\left(2ab\right)^2=\)

\(=\left(a^2-b^2\right)^2+\left(2ab\right)^2\)

\(A=\left(x^2+x+1\right)^2-4\left(x+2\right)^2+15\)

\(\Rightarrow A=\left(x^2+x+1\right)^2-\left[2\left(x+2\right)\right]^2+15\)

\(\Rightarrow A=\left(x^2+x+1+2x+2\right)\left(x^2+x+1-2x-2\right)+15\)

\(\Rightarrow A=\left(x^2+3x+3\right)\left(x^2-x-1\right)+15\)

\(\Rightarrow A=\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+3\right)\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1\right)+15\)

\(\Rightarrow A=\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(1\right)\)

Ta có : \(\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\\\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\\\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4},\forall x\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[\left(-\dfrac{3}{2}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[\left(\dfrac{1}{2}+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[4-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[4+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\dfrac{9}{4}+15\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{19}{4}.\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{27}{16}+15\left(x=-\dfrac{3}{2}\right)\\A\ge-\dfrac{95}{16}+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{267}{16}\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

\(\Rightarrow GTNN\left(A\right)=\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

`5.25.2.41.8`

`= 5.50.41.8`

`= 5.400.41`

`= 2000.41`

`= 82000`

Đặt \(n^2+4n+2013=p^2\left(p\in Z\right)\)

\(\Rightarrow n^2+4n+4+2009=p^2\)

\(\Rightarrow\left(n+2\right)^2+2009=p^2\)

\(\Rightarrow p^2-\left(n+2\right)^2=2009\)

\(\Rightarrow\left(p+n+2\right)\left(p-n-2\right)=2009\)

mà \(p+n+2>p-n-2\left(n\in N\right)\) và 2009 là số nguyên tố

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}p+n+2=2009\\p-n-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}p+n+2=-2009\\p-n-2=-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=1002\\p=1005\end{matrix}\right.\)

Vậy \(n=1002\) thỏa đề bài

A là số chính phương nên: \(A=n^2+n+6=k^2\)

\(\Rightarrow4n^2+4n+24=4k^2\)

\(\Rightarrow4n^2+4n+1+23=4k^2\)

\(\Rightarrow\left(2n+1\right)^2+23=4k^2\)

\(\Rightarrow4k^2-\left(2n+1\right)^2=23\)

\(\Rightarrow\left(2k-2n-1\right)\left(2k+2n+1\right)=23\)

Do \(k,n\in N\) nên: \(2k+2n+1>2k-2n-1\)

Ta có hệ:

\(\left\{{}\begin{matrix}2k+2n+1=23\\2k+2n+1=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2k+2n+1=23\\4k=24\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}12+2n+1=23\\k=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2n+13=23\\k=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2n=10\\k=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n=5\\k=6\end{matrix}\right.\)

Vậy: n=5

4S=1.2.3.4+2.3.4.4+3.4.5.4+...+k(k+1)(k+2).4=

=1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+k(k+1)(k+2)[(k+3)-(k-1)]=

=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-(k-1)k(k+1)(k+2)+k(k+1)(k+2)(k+3)=

=k(k+1)(k+2)(k+3)=k(k+3)(k+1)(k+2)=

=(k²+3k)(k²+3k+2)=(k²+3k)²+2(k²+3k)

=> 4S+1=(k²+3k)²+2(k²+3k)+1=[(k²+3k)+1]²

\(n^2+4n+2013=\left(n^2+4n+4\right)+2009=k^2\)

\(\Leftrightarrow\left(n+2\right)^2+2009=k^2\)

\(\Rightarrow\left(k-n-2\right)\left(k+n+2\right)=2009\)

\(\Rightarrow k-n-2\) và \(k+n+2\) là ước của 2009

Ta có các TH

\(\left\{{}\begin{matrix}k-n-2=-1\\k+n+2=-2009\end{matrix}\right.\) 

Hoặc

\(\left\{{}\begin{matrix}k-n-2=-2009\\k+n+2=-1\end{matrix}\right.\)

Hoặc

\(\left\{{}\begin{matrix}k-n-2=1\\k+n+2=2009\end{matrix}\right.\)

Hoặc

\(\left\{{}\begin{matrix}k-n-2=2009\\k+n+2=1\end{matrix}\right.\)

Giải các hệ trên tìm n

\(B=n^2-2.n.\dfrac{1}{2}+\dfrac{1}{4}+12,25=\)

\(=\left(n-\dfrac{1}{2}\right)^2+12,25\ge12,25\)

B là số chính phương

\(\Rightarrow n^2-n+13=p^2\) 

\(\Leftrightarrow4n^2-4n+52=4p^2\)

\(\Leftrightarrow\left(2n-1\right)^2+51=4p^2\)

\(\Leftrightarrow4p^2-\left(2n-1\right)^2=51\)

\(\Leftrightarrow\left(2p-2n+1\right)\left(2p+2n-1\right)=51\)

\(\Rightarrow\left(2p-2n+1\right)\) và \(\left(2p+2n-1\right)\) phải là ước của 51

\(=\left\{-51;-17;-3-1;1;3;17;51\right\}\)

Ta có các trường hợp

\(\left\{{}\begin{matrix}2p-2n+1=-51\\2p+2n-1=-1\end{matrix}\right.\) giải hệ để tìm n

Tương tự với các trường hợp khác

Bài trên:

\(16x^3y+0,25yz^3=\dfrac{1}{4}y\left(64x^3+z^3\right)=\dfrac{1}{4}y\left[\left(4x\right)^3+z^3\right]\\ =\dfrac{1}{4}y\left[\left(4x+z\right)\left(16x^2-4xz+z^2\right)\right]\\ ----\\ x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\\ -----\\ a^3+a^2b-ab^2-b^3=\left(a^3-b^3\right)+\left(a^2b-ab^2\right)\\ =\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)=\left(a-b\right)\left(a^2+2ab+b^2\right)=\left(a-b\right)\left(a+b\right)^2\)

Bài trên

\(x^3+x^2-4x-4\\ =x^2\left(x+1\right)-4\left(x+1\right)\\ =\left(x^2-4\right)\left(x+1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ ---\\ x^3-x^2-x+1\\ =x^2\left(x-1\right)-\left(x-1\right)\\ =\left(x^2-1\right)\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\\ ---\\ x^4+x^3+x^2-1\\ =x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\\ =\left(x^3+x-1\right)\left(x+1\right)\\ ---\\ x^2y^2+1-x^2-y^2\\ =x^2.\left(y^2-1\right)-\left(y^2-1\right)\\ =\left(y^2-1\right)\left(x^2-1\right)\\ =\left(y-1\right)\left(y+1\right)\left(x-1\right)\left(x+1\right)\)

X={9;10;11;12;...;78;79;80}

Phần tử nhỏ nhất: 9

Phần tử lớn nhất: 80

Khoảng cách giữa 2 phần tử liên tiếp: 10-9=1

b, Số lượng phần tử của tập hợp X:

(80-9):1 +1= 72 (phần tử)

72 phần tử