Question

gia tri lon nhat cua

2

___________________

x²-x+1

Answer 1

\(\frac{2}{x^2-x+1}=\frac{2}{x^2-x+\frac{1}{4}+\frac{3}{4}}=\frac{2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\)

Ta thấy: \(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow\frac{2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{2}{\frac{3}{4}}=\frac{8}{3}\)

Dấu "=" xảy ra khi \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Vậy \(Max=\frac{8}{3}\) khi \(x=\frac{1}{2}\)

Answer 2

Ta có:

\(\frac{2}{x^2-x+1}\)

\(=\frac{2}{x^2-2\cdot\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}}\)

\(=\frac{2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{2}{\frac{3}{4}}=\frac{8}{3}\)

Dấu bằng xảy ra khi \(x-\frac{1}{2}=0\)

\(x=0+\frac{1}{2}=\frac{1}{2}\)

Vậy GTLN của phân thức trên là \(\frac{8}{3}\) khi \(x=\frac{1}{2}\)