Toán Lớp 9: Tìm min √x ²-2x+5
( căn(x ² – 2x+5))
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Toán Lớp 9: Tìm min √x ²-2x+5
( căn(x ² – 2x+5))
Comments ( 1 )
$\sqrt{x^2-2x+5}$
$=\sqrt{x^2-2x+1+4}$
$=\sqrt{(x-1)^2+4}$
Ta có:
$(x-1)^2≥0∀x$
$⇔(x-1)^2+4≥4$
$⇔\sqrt{(x-1)^2+4}≥2$
Dấu $”=”$ xảy ra khi $x-1=0⇔x=1$
Vậy $Min=2⇔x=1$