a) \(A=4x^2-12x+10\)
\(A=4x^2-12x+9+1\)
\(A=\left(2x-3\right)^2+1\)
Vì \(\left(2x+3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow2x+3=0\Leftrightarrow x=-1,5\)
Vậy \(MIN_A=1\Leftrightarrow x=-1,5\)
b) \(B=3y^2+6y+5\)
\(B=3\left(y^2+2y+\dfrac{5}{3}\right)\)
\(B=3\left(y^2+2y+1+\dfrac{2}{3}\right)\)
\(B=3\left(y+1\right)^2+2\)
Vì \(3\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow3\left(y+1\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow y+1=0\Leftrightarrow y=-1\)
Vậy \(MIN_B=2\Leftrightarrow x=-1\)
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