-x² + 20x + 80

= -x² + 20x - 100 + 180

= -(x² - 20x  + 100) + 180

= 180 - (x - 10)² 

 \(=\left(\sqrt{180}\right)^2-\left(x-10\right)^2=\left(\sqrt{180}-x+10\right)\left(\sqrt{180}+x-10\right)\)

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 }

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

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

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

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

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

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

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

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

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

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

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

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