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\(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4\)
Kl: MinA = 4
\(B=x^2-x+1=\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
kl:.......
\(C=5x^2+5x+1=5\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+1-\dfrac{5}{4}=5\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
kl:.......
\(D=3x^2+4x+2=3\left(x^2+2\cdot\dfrac{2}{3}x+\dfrac{4}{9}\right)+2-\dfrac{4}{3}=3\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)
kl:......
\(E=\dfrac{1}{2}\cdot x^2+x-1=\dfrac{1}{2}\left(x^2+2x+1\right)-1-\dfrac{1}{2}=\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)
kl:............
\(F=\dfrac{1}{9}x^2+3x+2=\dfrac{1}{3}\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+2-\dfrac{1}{12}=\dfrac{1}{3}\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\)
kl:..........
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x2−x+14=0x2−x+14=0
⇔(x−12)2=0⇔(x−12)2=0
⇔x−12=0⇔x−12=0
⇔x=12⇔x=12
Vậy x=12x=12
Wish you study well !!
\(\frac{x^2-x+1}{4}=1\)
\(\Leftrightarrow x^2-x+1=4\)
\(\Leftrightarrow x^2-x-3=0\)
Đến đây lập delta giải phương trình bậc 2