\(M=19-6x-9x^2\)
\(-M=9x^2+6x-19\)
\(=\left(9x^2+6x+1\right)-20\)
\(=\left(3x+1\right)^2-20\)
\(Do\)\(\left(3x+1\right)^2\ge0\)\(\forall x\)
=>\(\left(3x+1\right)^2-20\ge-20\)\(\forall x\)
=>\(-M\ge-20\)\(\forall x\)
=> \(M\le20\)\(\forall x\)
Dấu = xảy ra khi:
\(\left(3x+1\right)^2=0\)
<=> \(3x+1=0\)
<=> \(3x=-1\)
<=> \(x=\frac{-1}{3}\)
Vậy \(M_{max}\)\(\le20\)\(khi\)\(x=\frac{-1}{3}\)
\(N=1+4x-x^2\)
\(-N=x^2-4x+1\)
\(=\left(x^2-4x+4\right)-3\)
\(=\left(x-2\right)^2-3\)
\(Do\)\(\left(x-2\right)^2\)\(\ge0\)\(\forall x\)
=>\(\left(x-2\right)^2-3\)\(\ge-3\)\(\forall x\)
=>\(-N\ge-3\)\(\forall x\)
=>\(N\le3\)\(\forall x\)
Dấu = xảy ra khi:
\(\left(x+2\right)^2=0\)
<=> \(x+2=0\)
<=>\(x=-2\)
Vậy \(N_{max}\)\(\le3\)\(khi\)\(x=-2\)
Chúc bạn học tốt ~! :)
+) \(M=19-6x-9x^2=-9x^2-6x+19=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)
Vì \(-\left(3x+1\right)^2\le0\Rightarrow M=-\left(3x+1\right)^2+20\le20\)
Dấu "=" xảy ra khi -(3x+1)2=0 <=>x=-1/3
Vậy Mmax=20 khi x=-1/3
+) \(N=1+4x-x^2=-x^2+4x+1=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\)
tiếp tục giống M
