\(\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)

\(=\sqrt{a\left(a+6\right)\left(a+1\right)\left(a+5\right)\left(a+2\right)\left(a+4\right)+36}\)

\(=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\left(1\right)\)

Đặt \(a^2+6a=x\), Ta có:

\(\left(1\right)=\sqrt{x\left(x+5\right)\left(x+8\right)+36}\)

\(=\sqrt{\left(x^2+5\right)\left(x+8\right)+36}=\sqrt{x^3+13x^2+40x+36}\)

\(=\sqrt{x^3+9x^2+4x^2+36x+4x+36}=\sqrt{\left(x+9\right)\left(x+2\right)^2}\)

Thay \(x=a^2+6a\)vào biểu thức trên ta được:

\(\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a^2+6a+2\right)^2}=\left(a+3\right)\left(a^2+6a+2\right)\)

\(\rightarrowđpcm\)

\(a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36=\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36\)
Đặt \(a^2+6a=t\) ta có:\(t\left(t+5\right)\left(t+8\right)+36=t\left(t^2+13t+40\right)=t^3+13t^2+40t+36=\left(t+9\right)\left(t+2\right)^2\)

Do đó \(\sqrt{\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}=\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a^2+6a+2\right)^2}\)

\(=\left(a+3\right)\left(a^2+6a+2\right)\)(Dấu () ở đây là giá trị tuyệt đối nha)

Do đó với a nguyên thì \(\left(a+3\right)\left(a^2+6a+2\right)\)nguyên (Dấu () ở đây là giá trị tuyệt đối nha) 

Vậy nếu a nguyên thì \(\sqrt{\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)nguyên

\(D=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\)

Đặt a^2+6a=x

=>\(D=\sqrt{x\left(x+5\right)\left(x+8\right)+36}\)

\(=\sqrt{x\left(x^2+13x+40\right)+36}\)

\(=\sqrt{x^3+13x^2+40x+36}\)

=>\(D=\sqrt{x^3+9x^2+4x^2+36x+4x+36}\)

\(=\sqrt{\left(x+9\right)\left(x^2+4x+4\right)}\)

\(=\sqrt{\left(a^2+6a+9\right)\left(x+2\right)^2}\)

=|a+3|*|x+2| là số nguyên

Nhìn cái D cồng kềnh thế thôi chứ key vô cùng EZ.

\(D=\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)

\(=\sqrt{\left[a\left(a+6\right)\right]\left[\left(a+1\right)\left(a+5\right)\right]\left[\left(a+2\right)\left(a+4\right)\right]+36}\)

\(=\sqrt{\left(a^2+6a\right)\left(a^2+6a+5\right)\left(a^2+6a+8\right)+36}\)

Đặt \(a^2+6a=x\)

Ta có:

\(D=\sqrt{x\left(x+5\right)\left(x+8\right)+36}=\sqrt{x^3+13x^2+40x+36}\)

\(=\sqrt{\left(x+9\right)\left(x+2\right)^2}\)

Thay \(x=a^2+6a\) ta có:

\(D=\sqrt{\left(a^2+6a+9\right)\left(a^2+6a+2\right)^2}=\sqrt{\left(a+3\right)^2\left(a+6a+2\right)^2}=\left(a+3\right)\left(a+6a+2\right)\)

là số nguyên vs a nguyên khác 0 nha !

\(\sqrt{a\left(a+1\right)\left(a+2\right)\left(a+4\right)\left(a+5\right)\left(a+6\right)+36}\)

=\(\sqrt{\left(a\left(a+4\right)\left(a+5\right)\right).\left(\left(a+1\right)\left(a+2\right)\left(a+6\right)\right)+36}\)

\(\sqrt{\left(a^3+9a^2+20a\right).\left(a^3+9a^2+20a+12\right)+36}\)

Đặt a^3+9a^2+20a+6=k(k thuộc Z)

ta có\(\sqrt{\left(k-6\right)\left(k+6\right)+36}=\sqrt{k^2-36+36}=\sqrt{k^2}=k\)

Vì k thuộc Z

=>A thuộc Z

tick nha

a =1  => A =2\(\sqrt{21}\)

CM đến sang năm

bỏ cái căn đi là chứng minh ngon lành ngay ^^

Đề hay thật sự, cho x,y,z nhưng chứng minh a,b,c :v

Ta có :

\(\left\{{}\begin{matrix}a+y^2=xy+yz+zx+y^2=\left(x+y\right)\left(y+z\right)\\a+z^2=xy+yz+zx+z^2=\left(x+z\right)\left(y+z\right)\\a+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\end{matrix}\right.\)

Do đó :

\(VT=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\left(x+z\right)\right)}}+y\sqrt{\dfrac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=2\left(xy+yz+zx\right)\)

\(=2a\) ( đpcm )