Ta có
\(x^2+y^2-2x-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1=\)
\(\left(x-1\right)^2+\left(y-2\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\) >0 => đpcm
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