Bài 6 :
a, \(A=x^2-2x+1+199=\left(x-1\right)^2+199\ge199>0\)
\(b,B=3\left(x^2-\dfrac{2}{3}x+2\right)=3\left(x^2-\dfrac{2.x.1}{3}+\dfrac{1}{9}+\dfrac{17}{9}\right)=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{17}{3}\ge\dfrac{17}{3}>0\)
\(c,C=x^2-2xy+y^2+y^2+2=\left(x-y\right)^2+y^2+2\ge2>0\)
b, \(M=-x^2+2x-1-1=-\left(x-1\right)^2-1\le-1< 0\)
\(N=-x^2+3x-10=-x^2+\dfrac{2.x.3}{2}-\dfrac{9}{4}-\dfrac{31}{4}=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{4}\le-\dfrac{31}{4}< 0\)
Bài 6 :
a, A=x2−2x+1+199=(x−1)2+199≥199>0A=x2−2x+1+199=(x−1)2+199≥199>0
N=−x2+3x−10=−x2+2.x.32−94−314=−(x−32)2−314≤−314<0
