Question

Cho biểu thức

P=\(\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

CMR P luôn âm với mọi x≠0 ,x≠ -3

Answer 1

ta có : \(P=\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow P=\dfrac{1-x^2}{x}\left(\dfrac{x^2-x-3}{x+3}\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow P=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)}{x^2+3x}+\dfrac{3x^2-14x+3}{x^2+3x}\) \(\Leftrightarrow P=\dfrac{x^2-x-3-x^4+x^3+3x^2}{x^2+3x}+\dfrac{3x^2-14x+3}{x^2+3x}\) \(\Leftrightarrow P=\dfrac{x^2-x-3-x^4+x^3+3x^2+3x^2-14x+3}{x^2+3x}\) \(\Leftrightarrow P=\dfrac{-x^4+x^3+7x^2-15x}{x^2+3x}\) \(\Leftrightarrow P=\dfrac{-x^4-3x^3+4x^3+12x^2-5x^2-15x}{x^2+3x}\)

\(\Leftrightarrow P=\dfrac{-x^2\left(x^2+3x\right)+4x\left(x^2+3x\right)-5\left(x^2+3x\right)}{x^2+3x}\)

\(\Leftrightarrow P=\dfrac{-\left(x^2-4x+5\right)\left(x^3+3x\right)}{x^2+3x}=-\left(x^2-4x+5\right)\)

\(\Leftrightarrow P=-\left(x^2-4x+4+1\right)=-\left(\left(x-2\right)^2+1\right)\)

\(\Leftrightarrow P=-\left(x-2\right)^2-1\le-1< 0\forall x\) (đpcm)