a: \(=\dfrac{x^2}{x-2}\cdot\dfrac{x^2+4-4x}{x}+3\)
\(=x\left(x-2\right)+3=x^2-2x+3\)
b: \(x^2-2x+3=\left(x-1\right)^2+2>=2\forall x\)
Dấu '=' xảy ra khi x=1
a,
\(\dfrac{x^2}{x-2}.\left(\dfrac{x^2+4}{x}-4\right)+3\)
\(=\dfrac{x^2}{x-2}.\left(\dfrac{x^2+4}{x}-\dfrac{4x}{x}\right)+3\)
\(=\dfrac{x^2}{x-2}\)\(.\)\(\dfrac{x^2+4-4x}{x}+3\)
\(=\dfrac{x^2}{x-2}.\)\(\dfrac{\left(x-2\right)^2}{x}+3\)
\(=\dfrac{x^2.\left(x-2\right)\left(x-2\right)}{\left(x-2\right).x}+3=x\left(x-2\right)+3\)
\(=x^2-2x+3\)
b, \(x^2-2x+3=x^2-2x+1+2=\left(x-1\right)^2+2\)
Ta luôn có : \(\left(x-1\right)^2\ge0\) với mọi x.
Suy ra : \(\left(x-1\right)^2+2\ge2\) với mọi x.
Dấu “=” xảy ra khi : x – 1 = 0 hay x = 1
Nên : Amin = 2 khi x = 1
