`a)A=x^2-10x-3=x^2-10x+25-28=(x-5)^2-28`
Vì `(x-5)^2 >= 0 AA x`
`<=>(x-5)^2-28 >= -28 AA x`
Hay `A >= -28 AA x`
Dấu "`=`" xảy ra `<=>(x-5)^2=0<=>x=5`
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`b)B=2x^2+4x+7=2(x^2+2x+1)+5=2(x+1)^2+5`
Vì `2(x+1)^2 >= 0 AA x`
`<=>2(x+1)^2+5 >= 5 AA x`
Hay `B >= 5 AA x`
Dấu "`=`" xảy ra `<=>(x+1)^2=0<=>x=-1`
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`c)C=3x^2-5x+1=3(x^2-5/3x+1/3)=3(x-5/6)^2-13/12`
Vì `3(x-5/6)^2 >= 0 AA x`
`<=>3(x-5/6)^2-13/12 >= -13/12 AA x`
Hay `C >= -13/12 AA x`
Dấu "`=`" xảy ra `<=>(x-5/6)^2=0<=>x=5/6`
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