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. \(\Rightarrowđpcm\)

a, Vì \(4-\sqrt{15}>0\) mà biểu thức \(f\left(x\right)=\left(4-\sqrt{15}\right)x+1\) là hàm số bậc nhất

⇒ Hàm số f(x) đồng biến.

b, \(f\left(0\right)=1\\ f\left(4+\sqrt{15}\right)=2\)

a) \(\left(x^2-25\right)^2-\left(x-5\right)^2\)

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

\(=\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]\)

\(=\left(x-5\right)^2\left(x^2+10x+25-1\right)\)

\(=\left(x-5\right)^2\left(x^2+10x+24\right)\)

b) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(2x-5\right)^2\left(2x+5\right)^2-9\left(2x-5\right)^2\)

\(=\left(2x-5\right)^2\left[\left(2x+5\right)^2-9\right]\)

\(=\left(2x-5\right)^2\left(4x^2+20x+25-9\right)\)

\(=\left(2x-5\right)^2\left(4x^2+20x+16\right)\)

c) \(4\left(2x-3\right)^2-9\left(4x^2-9\right)^2\)

\(=4\left(2x-3\right)^2-9\left(2x-3\right)^2\left(2x+3\right)^2\)

\(=\left(2x-3\right)^2\left[4-9\left(2x+3\right)^2\right]\)

\(=\left(2x-3\right)^2\left[4-9\left(4x^2+12x+9\right)\right]\)

\(=\left(2x-3\right)^2\left(4-36x^2-108x-81\right)\)

\(=(2x-3)^2(-36x^2-108x-77)\)

d) \(a^6-a^4+2a^3+2a^2\)

\(=\left(a^6-a^4\right)+\left(2a^3+2a^2\right)\)

\(=a^4\left(a^2-1\right)+2a^2\left(a-1\right)\)

\(=a^4\left(a+1\right)\left(a-1\right)+2a^2\left(a-1\right)\)

\(=\left(a-1\right)\left[a^4\left(a+1\right)+2a^2\right]\)

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

\(=a^2\left(a-1\right)\left(a^3+a^2+2\right)\)

e) \(\left(3x^2+3x+2\right)^2-\left(3x^2+2x-2\right)^2\)

\(=\left[\left(3x^2+3x+2\right)-\left(3x^2+3x-2\right)\right]\left[\left(3x^2+3x+2\right)+\left(3x^2+3x-2\right)\right]\)

\(=\left(3x^2+3x+2-3x^2-3x+2\right)\left(3x^2+3x+2+3x^2+3x-2\right)\)

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

\(=4\cdot6x\left(x+1\right)\)

\(=24x\left(x+1\right)\)

\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)

\(=\dfrac{x+1+x-18+x+2}{x-5}\)

\(=\dfrac{3x-15}{x-5}\)

\(=\dfrac{3\left(x-5\right)}{x-5}\)

\(=\dfrac{3}{1}\)

\(=3\)

\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

a) \(x^3+6x^2+12x+8\)

\(=x^3+3\cdot x^2\cdot2+3\cdot2^2\cdot x+2^3\)

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

b) \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

\(=\left(x-1\right)^3\)

c) \(1-9x+27x^2-27x^3\)

\(=1^3-3\cdot3x\cdot1^2+3\cdot\left(3x\right)^2\cdot1-\left(3x\right)^3\)

\(=\left(1-3x\right)^3\)

d) \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

\(=x^3\cdot3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3\)

\(=\left(x+\dfrac{1}{2}\right)^3\)

e) \(27x^3-54x^2y+36xy^2-8y^3\)

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

\(=\left(3x-2y\right)^3\)

d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left[\left(3x+1\right)-2\left(x-2\right)\right]\left[\left(3x+1\right)+2\left(x-2\right)\right]\)

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

\(=\left(x+5\right)\left(5x-3\right)\)

c) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left[3\left(2x+3\right)-2\left(x+1\right)\right]\left[3\left(2x+3\right)+2\left(x+1\right)\right]\)

\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)

\(=\left(4x-7\right)\left(8x+11\right)\)

f) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

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

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

g: =(ax+by-ay-bx)(ax+by+ay+bx)

=[a(x-y)-b(x-y)]*[a(x+y)+b(x+y)]

=(x-y)(x+y)(a-b)(a+b)

h: =(a^2+b^2-5-2ab-4)(a^2+b^2-5+2ab+4)

=[(a-b)^2-9][(a+b)^2-1]

=(a-b-3)(a-b+3)(a+b-1)(a+b+1)

i: =(4x^2-3x-18-4x^2-3x)(4x^2-3x-18+4x^2+3x)

=(-6x-18)(8x^2-18)

=-12(x+3)(4x^2-9)

=-12(x+3)(2x-3)(2x+3)

k: =(3x+3y-3)^2-(4x+6y+2)^2

=(3x+3y-3-4x-6y-2)(3x+3y-3+4x+6y+2)

=(-x-3y-5)(7x+9y-1)

i: =25-(2x-3y)^2

=(5-2x+3y)(5+2x-3y)

m: =(x-y)^2-(2m-n)^2

=(x-y-2m+n)(x-y+2m-n)

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

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

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

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

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

\(\Rightarrow dpcm\)

ko bt

\(x^2-25=\left(x-5\right)\left(x+5\right)---\\ 9x^2-\dfrac{1}{16}y=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2=\left(3x-\dfrac{1}{4}y\right)\left(3x+\dfrac{1}{4}y\right)\\ ----\\ x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2\\ =\left(x^3-y^2\right)\left(x^3+y^2\right)\\ ---\\ \left(2x-5\right)^2-64=\left(2x-5\right)^2-8^2\\ =\left(2x-5-8\right)\left(2x-5+8\right)=\left(2x-13\right)\left(2x+3\right)\)

\(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2\\ =\left(9+3x+2\right).\left(9-3x-2\right)=\left(11+3x\right)\left(7-3x\right)\\ ---\\ 9\left(x-5y\right)^2-16\left(x+y\right)^2\\ =9\left(x^2-10xy+25y^2\right)-16\left(x^2+2xy+y^2\right)\\ =9x^2-90xy+225y^2-16x^2-32xy-16y^2\\ =-7x^2-122xy+209y^2=-7x^2-133xy+11xy+209y^2\\ =-7x\left(x+19y\right)+11y\left(x+19y\right)\\ =\left(11y-7x\right)\left(x+19y\right)\)

Với x = y có vẻ ai cũng đồng ý, nhưng sẽ thế nào nếu x khác y? Em thử giải nhé, đây là đề chọn đội tuyển Anh môn Toán năm 2023!

\(x^y=y^x\)

\(\Rightarrow xy=yx\)

\(\Rightarrow xy:xy=xy:xy\)

\(\Rightarrow1=1\) (luôn đúng) 

Nên phương trình luôn đúng với mọi \(x=y\)

⇒ \(x,y\in R\)