a/ \(M=6x-x^2-5\)
\(=-\left(-6x+x^2+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left[\left(x-3\right)^2-4\right]\)
\(=-\left(x-3\right)^2+4\)
Vì \(-\left(x-3\right)^2\le0\) nên \(-\left(x-3\right)^2+4\le4\)
Vậy \(Max_M=4\) khi x = 3
b/ \(N=5-4x^2+4x\)
\(=-\left(-5+4x^2-4x\right)\)
\(=-\left(4x^2-4x+1-6\right)\)
\(=-\left[\left(2x-1\right)^2-6\right]\)
\(=-\left(2x-1\right)^2+6\le6\)
Vậy \(Max_N=6\) khi \(x=\dfrac{1}{2}\)
c/ \(P=-x^2-4x-y^2+2y\)
\(=-\left(x^2+4x+y^2-2y\right)\)
\(=-\left[\left(x^2+4x+4\right)+\left(y^2-2y+1\right)-5\right]\)
\(=-\left[\left(x+2\right)^2+\left(y-1\right)^2-5\right]\)
\(=-\left(x+2\right)^2-\left(y-1\right)^2+5\le5\)
Vậy \(Max_P=5\) khi x = -2 và y = 1
\(N=5-4x^2+4x=-\left(4x^2-4x+1\right)+6\)
\(=-\left(2x-1\right)^2+6\)
Vì \(\left(2x-1\right)^2\ge0\forall x\Rightarrow-\left(2x-1\right)^2\le0\forall x\Rightarrow-\left(2x-1\right)^2+6\le6\forall x\)
\(\Rightarrow N_{max}=6\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy, N đạt GTLN là 6 <=> \(x=\dfrac{1}{2}\)
\(M=6x-x^2-5=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9\right)+4=-\left(x-3\right)^2+4\)
Vì \(\left(x-3\right)^2\ge0\forall x\Rightarrow-\left(x-3\right)^2\le0\forall x\Rightarrow-\left(x-3\right)^2+4\le4\forall x\)
\(\Rightarrow M_{max}=4\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy, M đạt GTLN là 4 <=> x = 3