Bài làm

Ta có : \(\frac{x-2}{x+3}=\frac{x+3-5}{x+3}=1-\frac{5}{x+3}\) ( x khác -3 )

Để biểu thức có giá trị nguyên thì \(\frac{5}{x+3}\)đạt giá trị nguyên

=> 5 chia hết cho ( x + 3 )

=> ( x + 3 ) thuộc Ư(5) = { ±1 ; ±5 }

Các giá trị trên đều thỏa mãn ĐKXĐ

Vậy x thuộc { -8 ; -4 ; -2 ; 2 }

\(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right)\div\frac{6x^2+10x}{1-6x+9x^2}\)

\(=\left(\frac{-3x}{3x-1}+\frac{2x}{3x+1}\right)\div\frac{2x\left(3x+5\right)}{9x^2-6x+1}\)

\(=\left(\frac{-3x\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}+\frac{2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\right)\div\frac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)

\(=\left(\frac{-9x^2-3x}{\left(3x-1\right)\left(3x+1\right)}+\frac{6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}\right)\div\frac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)

\(=\frac{-3x^2-5x}{\left(3x-1\right)\left(3x+1\right)}\times\frac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=\frac{-x\left(3x+5\right)\times\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)\times2x\left(3x+5\right)}\)

\(=\frac{-\left(3x-1\right)}{2\left(3x+1\right)}=\frac{-3x+1}{6x+2}\)

\(=\frac{\left(x^2+1\right)\left(x^8+2x^4+1-x^4\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)}{x^4+x^2+1}\)

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

\(\frac{\left(x^2+1\right)\left(x^8+x^4+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{\left(x^2+1\right)\left(x^8+2x^4+1-x^4\right)}{\left(x^2+1\right)^2-x^2}\)

\(=\frac{\left(x^2+1\right)\left[\left(x^4+1\right)^2-x^4\right]}{x^4+2x^2+1-x^2}=\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)}{x^4+x^2+1}\)

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

\(\frac{3x}{2x+4};\frac{x+3}{x^2-4}\)

Ta ó : \(2x+4=2\left(x+2\right)\)

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

MTC : \(2\left(x-2\right)\left(x+2\right)\)

\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{3x^2-6x}{2\left(x+2\right)\left(x-2\right)}\)

\(\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x+3\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{2x+6}{2\left(x-2\right)\left(x+2\right)}\)

\(\hept{\begin{cases}\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}\\\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}\end{cases}}\)

MTC : 2( x - 2 )( x + 2 )

=> \(\hept{\begin{cases}\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x}{2\left(x-2\right)\left(x+2\right)}\\\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x+3\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{2x+6}{2\left(x-2\right)\left(x+2\right)}\end{cases}}\)

Đặt x + 2 = t ta có  biểu thứ mới : 

\(t\left(t+1\right)^2\left(t+2\right)-12=t\left(t^2+2t+1\right)\left(t+2\right)-12\)

\(=\left(t^3+2t^2+t\right)\left(t+2\right)-12=t^4+2t^3+2t^3+4t^2+t^2+2t-12\)

\(=t^4+4t^3+5t^2+2t-12=\left(t-1\right)\left(t^3+5t^2+10t+12\right)\)

\(=\left(t-1\right)\left(t+3\right)\left(t^2+2t+4\right)=\left(x+1\right)\left(x+5\right)\left[\left(x+2\right)^2+2\left(x+2\right)+4\right]\)

( x + 2 )( x + 3 )²( x + 4 ) - 12

= [ ( x + 2 )( x + 4 ) ]( x + 3 )² - 12

= ( x² + 6x + 8 )( x² + 6x + 9 ) - 12

Đặt t = x² + 6x + 8

= t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + 6x + 8 - 3 )( x² + 6x + 8 + 4 )

= ( x² + 6x + 5 )( x² + 6x + 12 )

= ( x² + 5x + x + 5 )( x² + 6x + 12 )

= [ x( x + 5 ) + ( x + 5 ) ]( x² + 6x + 12 )

= ( x + 5 )( x + 1 )( x² + 6x + 12 )

\(a^7-b^7=\left(a-b\right)\left(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6\right)\)

\(a^7-b^7=\left(a-b\right)\left(a^6+a^5b+a^4b^2+a^3b^c+a^2b^4+ab^5+b^6\right)\)

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow x+y+z=0\).

\(P=\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz}{\left(-x\right)\left(-y\right)\left(-z\right)}=-1\)