Đáp án + Giải thích các bước giải:
$A=x^{2}-6x+10=(x^{2}-6x+9)+1=(x-3)^{2}+1\geq1$ ∀$x$
$($ Vì $(x-3)^{2}\geq$0 ∀$x$$)$
$⇒A\geq1$
Dấu "$=$" xảy ra $⇔(x-3)^{2}=0⇔x-3=0⇔x=3$
Vậy $GTNN$ của $A=1$ đạt khi $x=3$
Đáp án:
`min_A=1<=>x=3.`
Giải thích các bước giải:
`A=x^2-6x+10`
`A=x^2-3x-3x+9+1`
`A=x(x-3)-3(x-3)+1`
`A=(x-3)(x-3)+1`
`A=(x-3)^2+1`
Vì `(x-3)^2>=0`
`=>A>=1`
Dấu "= "xảy ra khi `x-3=0<=>x=3`
Vậy `min_A=1<=>x=3.`
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