Question

Chứng minh rằng:

a) x² - 2xy + 6y² - 12x + 2y + 41 (lớn hơn hoặc bằng) 0

b) \(\dfrac{x^2}{y^2}\) + \(\dfrac{y^2}{x^2}\) - \(\dfrac{2x}{y}\) - \(\dfrac{2y}{x}\) + 3 > 0

Answer 1

\(a.x^2-2xy+6y^2-12x+2y+41\)

\(=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5\)

\(=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y-1\right)^2\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2\) ≥ \(0\)

\(b.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-\dfrac{2x}{y}-\dfrac{2y}{x}+3\)

\(=\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1+1\)

\(=\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2+1>0\)