a) Ta có: \(A=x^2-5x+7\)
\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
b) Ta có: \(B=2x^2-8x+15\)
\(=2\left(x^2-4x+\dfrac{15}{2}\right)\)
\(=2\left(x^2-4x+4+\dfrac{7}{2}\right)\)
\(=2\left(x-2\right)^2+7\ge7\forall x\)
Dấu '=' xảy ra khi x=2
a. `A=x^2-5x+7`
`=x^2-2.x. 5/2 + (5/2)^2 +3/4`
`=(x-5/2)^2 + 3/4`
`=> A_(min) =3/4 <=> x-5/2 =0<=>x=5/2`
b) `B=2x^2-8x+15`
`=[(\sqrt2x)^2 -2.\sqrt2 x . 2\sqrt2 +(2\sqrt2)^2] +7`
`=(\sqrt2x-2\sqrt2)^2+7`
`=> B_(min)=7 <=> x=2`.
a) \(A=x^2-5x+7\)
\(=x^2-2.\dfrac{5}{2}x+\left(\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
Mặt khác, ta có \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\) \(\Rightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra khi \(\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(A_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{5}{2}\)
b) \(B=2x^2-8x+15\)
\(=4x^2-2.2x.2+2^2+11\)
\(=\left(2x-2\right)^2+11\)
Vì \(\left(2x-2\right)^2\ge0\forall x\) nên \(\left(2x-2\right)^2+11\ge11\forall x\)
Dấu "=" xảy ra khi \(\left(2x-2\right)^2=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
Vậy \(B_{min}=11\) khi \(x=1\)
