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Q(x) có nghiệm <=>Q(x)=0

=>2x^2-2x+10=0

can't solve

a) \(A=2x^2-6x=2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\forall x\)

Vậy GTNN của A = -9/2 khi x = 3/2.

b) \(B=x^2-x+\frac{1}{4}+y^2+6y+9+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x;y\)

Vậy, GTNN của B = 3/4 khi x=1/2 và y=-3

\(A=\dfrac{2x+1+4}{2x+1}=1+\dfrac{4}{2x+1}\)

A min khi 2x+1=-1

=>x=-1

\(2^{x+3}.2=2^2.3+52\)

\(=>2^{x+3}.2=64\)

\(=>2^{x+3}=64:2\)

\(=>2^{x+3}=32\)

\(=>2^{x+3}=2^5\)

=>x+3=5

=>x=5-3

=>x=2

Vậy ...........

 2^{x + 3} . 2 = 2² . 3 + 52

 2^{x + 3} . 2 = 4 . 3 + 52

 2^{x + 3} . 2 = 12 + 52

 2^{x + 3} . 2 = 64

     2^{x + 3}  = 64 : 2

     2^{x + 3}  = 32

     2^{x + 3} = 2⁵

     x + 3 = 5

           x = 5 - 3

           x = 2

Vậy x = 2

Ta có: \(A=\left|x-1999\right|+\left|x-9\right|=\left|1999-x\right|+\left|x-9\right|\ge\left|1999-x+x-9\right|=1990\)

Dấu "=" xảy ra khi \(\left(1999-x\right)\left(x-9\right)\ge0\Leftrightarrow9\le x\le1999\)

Vậy MinA = 1990 khi \(9\le x\le1999\)

\(A=\dfrac{x-3-2}{x-3}=1-\dfrac{2}{x-3}\)

A max khi -2/x-3 max

=>2/x-3 min

=>x-3=-1

=>x=2