Question

[Lê Minh Đức]

Tính: \(\sqrt{2012-2\sqrt{2011}}+1\)

CMR: \(\frac{1}{3}< =\frac{x^2+x+1}{x^2-x+1}< =3\)

Answer 1

[Hoàng Lê Bảo Ngọc]

• \(\sqrt{2012-2\sqrt{2011}}+1=\sqrt{\left(\sqrt{2011}-1\right)^2}+1=\sqrt{2011}\)
•  

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

\(\frac{x^2+x+1}{x^2-x+1}=\frac{3\left(x^2-x+1\right)-2\left(x^2-2x+1\right)}{x^2-x+1}=-\frac{2\left(x-1\right)^2}{x^2-x+1}+3\le3\)(2)

Từ (1) và (2) suy ra đpcm

Answer 2

[Lê Minh Đức]

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