\(a.P=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-32\)
\(P=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-32\)
Đặt : \(x^2+5x+5=t\) , ta có :
\(\left(t-1\right)\left(t+1\right)-32=t^2-1-32=t^2-33=\left(t-\sqrt{33}\right)\left(t+\sqrt{33}\right)\)
Thay : \(x^2+5x+5=t\) , ta có :
\(\left(x^2+5x+5-\sqrt{33}\right)\left(x^2+5x+5+\sqrt{33}\right)\)
\(b.Q=x^2-2xy+y^2+3x-3y+1=\left(x-y\right)^2-3\left(x-y\right)+1=\left(x-y\right)^2-2.\dfrac{3}{2}\left(x-y\right)+\dfrac{9}{4}+1-\dfrac{9}{4}=\left(x-y-\dfrac{3}{2}\right)^2-\dfrac{5}{4}=\left(x-y-\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)\left(x-y-\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)=\left(x-y-\dfrac{3+\sqrt{5}}{2}\right)\left(x-y+\dfrac{\sqrt{5}-3}{2}\right)\)
\(c.R=4x^2+\dfrac{1}{x^2}-20=4x^2-2.2x.\dfrac{1}{x}+\dfrac{1}{x^2}-16=\left(2x-\dfrac{1}{x}\right)^2-16=\left(2x-\dfrac{1}{x}-4\right)\left(2x-\dfrac{1}{x}+4\right)=\left(\dfrac{2x^2-1}{x}-4\right)\left(\dfrac{2x^2-1}{x}+4\right)\)
