a: \(A=x^2+3x+\dfrac{9}{4}+y^2-6y+9+1993.75\)
\(=\left(x+\dfrac{3}{2}\right)^2+\left(y-3\right)^2+1993.75>=1993.75\)
Dấu '=' xảy ra khi x=-3/2 và y=3
b: \(=3\left(x^2+\dfrac{7}{3}x+3\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{59}{36}\right)\)
\(=3\left(x+\dfrac{7}{6}\right)^2+\dfrac{59}{12}>=\dfrac{59}{12}\)
Dấu '=' xảy ra khi x=-7/6
c: \(=4\left(x^2-\dfrac{15}{4}x+5\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{15}{8}+\dfrac{225}{64}+\dfrac{95}{64}\right)\)
\(=4\left(x-\dfrac{15}{8}\right)^2+\dfrac{95}{16}>=\dfrac{95}{16}\)
Dấu '=' xảy ra khi x=15/8
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