a.
\(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
\(\left(x-1\right)^2\ge0\)
\(\left(x-1\right)^2+4\ge4\)
Vậy Min P = 4 khi x = 1
b.
\(Q=2x^2-6x=2\left(x^2-3x\right)=2\left[x^2-2\times x\times\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right]=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\)
\(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
\(2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\ge-\frac{9}{2}\)
Vậy Min Q = \(-\frac{9}{2}\) khi x = \(\frac{3}{2}\)
a)P=x2-2x+5
Ta có:P=x2-2x+5
P=x2-2x+1+4
P=(x-1)2+4
Vì (x-1)2\(\ge\)0
Suy ra:(x-1)2+4\(\ge\)4
Dấu = xảy ra khi x-1=0
x=1
Vậy Min P=4 khi x=1
b)Q=2x2-6x
Ta có:Q=2x2-6x
Q=2.(x2-2.1,5x+2,25)-4,5
Q=2.(x-1,5)2-4,5
Vì 2.(x-1,5)2\(\ge\)0
Suy ra:2.(x-1,5)2-4,5\(\ge\)-4,5
Dấu = xảy ra khi x-1,5=0
x=1,5
Vậy Min Q=-4,5 khi x=1,5
