\(A=\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(x+\dfrac{5}{2}\right)\ge;\forall x\)
\(\Rightarrow A\ge\dfrac{3}{4};\forall x\)
\(A_{min}=\dfrac{3}{4}\) khi \(x+\dfrac{5}{2}=0\Rightarrow x=-\dfrac{5}{2}\)
\(B=2\left(x^2-6x\right)=2\left(x^2-6x+9-9\right)=2\left(x^2-6x+9\right)-18=2\left(x-3\right)^2-18\)
Do \(2\left(x-3\right)^2\ge0;\forall x\)
\(\Rightarrow B\ge-18\)
\(B_{min}=-18\) khi \(x-3=0\Rightarrow x=3\)
`A = x^2 + 5x + 7 `
`= x^2 + 2 . x . 5/2 + (5/2)^2 + 3/4`
`= (x + 5/2)^2 + 3/4`
Do `(x + 5/2)^2 >= 0`
`=> (x + 5/2)^2 + 3/4 >= 3/4`
Dấu = có khi:
`x + 5/2 = 0`
`<=> x = -5/2`
Vậy ...
`B = 2x^2 - 12x `
`= 2(x^2 - 6x) `
`= 2(x^2 - 6x + 9 - 9)`
`= 2(x - 3)^2 - 18`
Do `(x - 3)^2 >= 0 => 2(x - 3)^2 >=0`
`=> 2(x - 3)^2 - 18 >=-18`
Dấu = có khi:
`x - 3 = 0`
`<=> x = 3`
Vậy ...
