Question

 

 Chứng minh rằng:

\(x^2\) +\(5y^2\) +2x - 4xy - 10y + 14 > 0 với mọi x, y.

Answer 1

Ta có : \(x^2+5y^2+2x-4xy-10y+14\)

\(=x^2+2x\left(1-2y\right)+\left(1-2y\right)^2-\left(1-2y\right)^2+5y^2-10y+14\)

\(=\left(x-2x+1\right)^2-1-4y^2+4y+5y^2-10y+14\)

\(=\left(x-2x+1\right)^2+y^2-6y+9+4\)

\(=\left(x-2x+1\right)^2+\left(y-3\right)^2+4\ge4>0\) (đpcm)

Answer 2

Ta có: x² + 5y² + 2x - 4xy - 10y + 14 

= (x² - 4xy + 4y²) + (2x - 4y) + 1 + (y² - 6y + 9) + 4

= (x - 2y)² + 2(x - 2y) + 1 + (y - 3)² + 4

= (x - 2y + 1)² + (y - 3)² + 4 > 0 \(\forall\)x; y

Do (x - 2y + 1)² \(\ge\)0; (y - 3)² \(\ge\)0 ; 4 > 0

Answer 3

\(x^2+5y^2+2x-4xy+10y+14\)

\(=\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+y^2-6y+13\)

\(=\left(x+1-2y\right)^2+\left(y^2-2y\cdot3+9\right)+4\)

\(=\left(x+1-2y\right)^2+\left(y-3\right)^2+4\)

Ta có: \(\hept{\begin{cases}\left(x+1-2y\right)^2\ge0\forall x\inℝ\\\left(y-3\right)^2\ge0\forall x\inℝ\end{cases}}\)

=> \(\left(x+1+2y\right)^2+\left(y-3\right)^2+4\ge4\)

=> \(x^2+5y^2+2x-4xy-10y+14>0\)(đpcm)