A=2(x^2-10x+25)+53-50
A=2(x-5)^2+3>=3
GTnnA=3 khi x=5
\(B=2x^2+3x+1=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)
\(=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{1}{16}\right)=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
Suy ra: \(Min_B=-\dfrac{1}{8}\) khi \(x=-\dfrac{3}{4}\)
A = 2x2 - 20x + 53
= \([\left(x\sqrt{2}\right)^2-2.x\sqrt{2}.5\sqrt{2}+\left(5\sqrt{2}\right)^{2^{ }}]+3\)
= \(\left(x\sqrt{2}-5\sqrt{2}\right)^{2^{ }}+3\ge3\)
Dấu "=" xảy ra khi \(x\sqrt{2}-5\sqrt{2}=0\) <=> \(x\sqrt{2}=5\sqrt{2}\) <=> x=5
Vậy MinA = 3 khi x = 5
B = 2x2 + 3x + 1
= \([\left(x\sqrt{2}\right)^{2^{ }}+2.x\sqrt{2}.\dfrac{3\sqrt{2}}{4}+\left(\dfrac{3\sqrt{2}}{4}\right)^{2^{ }}]-\dfrac{1}{8}\)
= \(\left(x\sqrt{2}+\dfrac{3\sqrt{2}}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
Dấu "=" xảy ra khi \(x\sqrt{2}+\dfrac{3\sqrt{2}}{4}=0\) <=> \(x\sqrt{2}=\dfrac{-3\sqrt{2}}{4}\) <=> x = \(\dfrac{-3}{4}\)
Vậy MinB = \(\dfrac{-1}{8}\) khi x = \(\dfrac{-3}{4}\)
\(A=2x^2-20x+53=2\left(x^2-10x+\dfrac{53}{2}\right)\)
\(A=2\left(x^2-2.x.5+25+\dfrac{3}{2}\right)\)
\(A=2\left(x-5\right)^2+3\ge3\)
Suy ra: \(Min_A=3\) khi \(x=5\)
8B=16x^2+3.8x+8
8B=[(4x)^2+2.3.(4x)+9]-1
8B=(4x+3)^2-1>=-1
B=>-1/8
khi x=-3/4