Question

Tìm giá trị nhỏ nhất

1) y = x²+ 5x - 4

2) y = 2x² - 6x + 5

3) y = x² - 4x + 1

4) 2x² - 8x + 3

5) y = \(\frac{2x+1}{x^2}\)

6) y = \(\frac{4x^2-2x+1}{x^2}\)

Answer 1

1)

\(y=x^2+5x-4\)

\(=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\left(\frac{25}{4}+4\right)\)

\(=\left(x+\frac{5}{2}\right)^2-10,25\)

\(Min_y=-10,25\Leftrightarrow x=-\frac{5}{2}\)

Answer 2

2) \(y=2x^2-6x+5\)

\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{1}{2}\)

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

\(Min_y=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)

Answer 3

3) \(y=x^2-4x+1\)

\(=\left(x^2-4x+4\right)-3\)

\(=\left(x-2\right)^2-3\)

\(Min_y=-3\Leftrightarrow x=2\)

Answer 4

4)

\(y=2x^2-8x+3\)

\(=2\left(x^2-4x+4\right)-5\)

\(=2\left(x-2\right)^2-5\)

\(Min_y=-5\Leftrightarrow x=2\)

Answer 5

5)

\(y=\frac{2x+1}{x^2}\)

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

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

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

\(Min_y=-1\Leftrightarrow x+1=0;x=-1\)

Answer 6

6)\(y=\frac{4x^2-2x+1}{x^2}\)

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

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

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

\(Min_y=3\Leftrightarrow x-1=0;x=1\)