a ) \(A=x^2-4x-7\)
\(A=\left(x^2+2.x.2+2^2\right)-11\)
\(A=\left(x+2\right)^2-11\)
Ta có : \(\left(x+2\right)^2\ge0\)
\(\Rightarrow\left(x+2\right)^2-11\ge-11\)
Vậy GTNN của \(A=-11\)
Khi : \(x+2=0\)
\(x=-2\)
b ) \(B=-x^2+4x-7\)
\(B=-\left(x^2+2.x.2-2^2\right)-3\)
\(B=-\left(x-2\right)^2-3\)
Ta có : \(-\left(x-2\right)^2\le0\)
\(\Rightarrow-\left(x-2\right)^2-3\le-3\)
Vậy GTLN của \(B=-3\)
Khi \(x-2=0\)
\(x=2\)
a)
\(A=\left(x^2-4x+4\right)-11\)
\(=\left(x-2\right)^2-11\)
Ta có
\(\left(x-2\right)^2-11\ge-11\)
Dấu " = " xảy ra khi x = 2
Vậy MINA= - 11 khi x=2
b)
\(B=-\left(x^2-4x+4\right)-3\)
\(B=-\left(x-2\right)^2-3\)
Ta có
\(-\left(x-2\right)^2-3\le-3\) với mọi x
Dấu " = " xảy ra khi = 2
Vậy MAXB= - 3 khi x = 2
a)\(A=x^2-4x-7\)
Ta có:\(x^2-4x-7=-\left(x^2+4x+7\right)\)
\(=-\left(x^2+2.2x+2^2\right)-3\)
\(=-\left(x+2\right)^2-3\)
Vì \(-\left(x+2\right)^2\le0\)
Suy ra:\(-\left(x+2\right)^2-3\le-3\)
b)\(B=-x^2+4x-7\)
Ta có:\(-x^2+4x-7=-\left(x^2-4x+7\right)\)
\(=-\left(x^2-2.2x+2^2\right)-3\)
\(=-\left(x-2\right)^2-3\)
Dấu = xảy ra khi x-2=0
x=2
Vậy MinA=-3 khi x=2
a) \(A=x^2-4x-7=\left(x^2-4x+4\right)-11=\left(x-2\right)^2-11\ge-11\)
Min A = -11 <=> x = 2
b) \(B=-x^2+4x-7=-\left(x^2-4x+4\right)-3=-\left(x-2\right)^2-3\le-3\)
Max B = -3 <=> x = 2