a,
\(x^2+2\sqrt{5}x+5=x^2+\sqrt{5}x+\sqrt{5}x+5=x\left(x+\sqrt{5}\right)+\sqrt{5}\left(x+\sqrt{5}\right)=\left(x+\sqrt{5}\right)^2\)
b,
Làm quen câu a rồi, bây giờ dùng HĐT
\(2x^2+2\sqrt{2}x+1=\left(\sqrt{2}x\right)^2-2\sqrt{2}x\cdot1+1^2=\left(\sqrt{2}x-1\right)^2\)
c,
\(x^2-2\sqrt{6}x+5=x^2-2\sqrt{6}x+\left(\sqrt{6}\right)^2-1=\left(x-\sqrt{6}\right)^2-1^2=\left(x-\sqrt{6}+1\right)\left(x-\sqrt{6}-1\right)\)
a) \(x^2+2\sqrt{5}x+5\)
\(=x^2+2\sqrt{5}x+\left(\sqrt{5}\right)^2\)
\(=\left(x+\sqrt{5}\right)^2\)
b) \(2x^2-2\sqrt{2}x+1\)
\(=2\left(x^2-\sqrt{2}x+\dfrac{1}{2}\right)\)
\(=2\left[x^2-2.x.\sqrt{2}.\dfrac{1}{2}+\left(\dfrac{\sqrt{2}}{2}\right)^2-\left(\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{1}{2}\right]\)
\(=2\left[x^2-2.x\sqrt{2}.\dfrac{1}{2}+\left(\dfrac{\sqrt{2}}{2}\right)^2-\dfrac{1}{2}+\dfrac{1}{2}\right]\)
\(=2\left(x-\dfrac{\sqrt{2}}{2}\right)^2\)
c) \(x^2-2\sqrt{6}x+5\)
\(=x^2-2\sqrt{6}x+\left(\sqrt{6}\right)^2-\left(\sqrt{6}\right)^2+5\)
\(=\left(x-\sqrt{6}\right)^2-6+5\)
\(=\left(x-\sqrt{6}\right)^2-1\)
\(=\left(x-\sqrt{6}-1\right)\left(x-\sqrt{6}+1\right)\)
