Question

Giúp mình với, cảm ơn bạn!

Bài 1. Tìm Max:

a. \(\frac{1}{2x-\sqrt{x}+1}\)

b.\(\frac{1}{x-2\sqrt{x}+3}\)

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

Bài 2. Tìm Min:

a.\(\frac{2}{6x-5-9x^2}\)

b.\(\frac{3x^2-8x+6}{x^2-2x+1}\)

Answer 1

a. \(2x-\sqrt{x}+1=2\left(\sqrt{x}-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

\(\Rightarrow\frac{1}{2x-\sqrt{x}+1}\le\frac{1}{\frac{7}{8}}=\frac{8}{7}\)

b.

\(x-2\sqrt{x}+3=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\frac{1}{x-2\sqrt{x}+3}\le\frac{1}{2}\)

c.

\(\sqrt{1-x^2}\ge0\Rightarrow1+\sqrt{1-x^2}\ge1\)

\(\Rightarrow\frac{1}{1+\sqrt{1-x^2}}\le1\)

2.

a. \(\frac{2}{6x-5-9x^2}=\frac{2}{-\left(3x-1\right)^2-4}\ge\frac{2}{-4}=-\frac{1}{2}\)

b. \(\frac{3x^2-8x+6}{x^2-2x+1}=\frac{2\left(x^2-2x+1\right)+x^2-4x+4}{x^2-2x+1}=2+\left(\frac{x-2}{x-1}\right)^2\ge2\)