\(M=19-6x-9x^2\)

\(-M=9x^2+6x-19\)

\(=\left(9x^2+6x+1\right)-20\)

\(=\left(3x+1\right)^2-20\)

\(Do\)\(\left(3x+1\right)^2\ge0\)\(\forall x\)

=>\(\left(3x+1\right)^2-20\ge-20\)\(\forall x\)

=>\(-M\ge-20\)\(\forall x\)

=> \(M\le20\)\(\forall x\)

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

+) \(E=x^2-6x+9+x^2-22x+121=2x^2-28x+130\)

\(\Rightarrow2E=4x^2-56x+242=\left(4x^2-56x+196\right)+46=\left(2x-14\right)^2+46\)

Vì \(\left(2x-14\right)^2\ge0\Rightarrow2E=\left(2x-14\right)^2+46\ge46\Rightarrow E\ge23\)

Dấu "=" xảy ra khi x=7 

Vậy Emin=23 khi x=7

+) \(F=\frac{-2}{x^2-2x+5}=\frac{-2}{x^2-2x+1+4}=\frac{-2}{\left(x-1\right)^2+4}\)

Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow F=\frac{-2}{\left(x-1\right)^2+4}\le-\frac{2}{4}=-\frac{1}{2}\)

Dấu "=" xảy ra khi x=1

Vậy Fmin=-1/2 khi x=1

+) \(G=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-6x+x-6\right)\left(x^2-3x-2x+6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)

Đặt x²-5x=t, ta được:

\(G=\left(t-6\right)\left(t+6\right)=t^2-36=\left(x^2-5x\right)^2-36\)

Vì \(\left(x^2-5x\right)^2\ge0\Rightarrow G=\left(x^2-5x\right)^2-36\ge36\)

Dấu "=" xảy ra khi x=0 hoặc x=5

Vậy Gmin=36 khi x=0 hoặc x=5

\(C=x^2-x+1=x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Nên GTNN của C là \(\frac{3}{4}\) đặt được khi \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

\(D=25x^2+3y^2-10xy+4y+1\)

\(=\left(5x\right)^2-2.5x.y+y^2+2y^2+4y+2-1\)

\(=\left(5x-y\right)^2+2\left(y+1\right)^2-1\ge-1\)

Nên GTNN của D là - 1  đạt được khi \(\hept{\begin{cases}y+1=0\\5x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-1\\y=5x\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-1\\x=-\frac{1}{5}\end{cases}}\)

\(A=x^2-4x+5\)

=\(\left(x^2-4x+4\right)+1\)

\(=\left(x+2\right)^2+1\)

Do \(\left(x+2\right)^2\ge0\forall x\)

=>\(\left(x+2\right)^2+1\ge1\forall x\)

=> \(A\ge1\forall x\)

Dấu = xảy ra khi:

\(\left(x+2\right)^2=0\)

<=> \(x+2=0\)

<=>\(x=-2\)

Vậy Amin \(\ge\) 1 khi \(x=-2\)

\(B=2x^2+4x+5\)

\(=\left(x^2+2x+1\right)+\left(x^2+2x+1\right)+3\)

\(=\left(x+1\right)^2+\left(x+1\right)^2+3\)

Do \(\left(x+1\right)^2\ge0\forall x\)

=>\(\left(x+1\right)^2+\left(x+1\right)^2+3\ge3\forall x\)

=> \(B\ge3\forall x\)

Dấu = xảy ra khi:

\(\left(x+1\right)^2=0\)

<=>\(x+1=0\)

<=> \(x=-1\)

Vậy  \(B_{min}\) \(\ge3\)\(khi\)\(x=-1\)

Chúc bạn học tốt~!

a,\(M=-2x^2+2x-3\)

\(\Rightarrow2M=-4x^2+4x-6=-\left(4x^2-4x+1\right)-5=-\left(2x-1\right)^2-5\)

Vì\(-\left(2x-1\right)^2\le0\Rightarrow2M=-\left(2x-1\right)^2-5\le-5\Rightarrow M\le-\frac{5}{2}\)

Dấu "=" xảy ra khi x=1/2

Vậy Mmax=-5/2 khi x=1/2

b, \(N=3x-x^2-4=-x^2+3x-4=-\left(x^2-3x+\frac{9}{4}\right)-\frac{7}{4}=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\)

Vì \(-\left(x-\frac{3}{2}\right)^2\le0\Rightarrow N=-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)

Dấu "=" xảy ra khi x=3/2

Vậy Nmax=-7/4 khi x=3/2

c, \(P=\frac{3}{x^2-6x+10}=\frac{3}{x^2-6x+9+1}=\frac{3}{\left(x-3\right)^2+1}\)

Vì \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1\Rightarrow\frac{1}{\left(x-3\right)^2+1}\le1\Rightarrow\frac{3}{\left(x-3\right)^2+1}\le3\)

Dấu "=" xảy ra khi x=3

Vậy Pmax=3 khi x=3

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a, \(A=x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi x=-1/2

Vậy Amin=3/4 khi x=-1/2

b,\(B=2x^2-5x-2\)

\(\Rightarrow2B=4x^2-10x-4=\left(4x^2-10x+\frac{25}{4}\right)-\frac{41}{4}=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\)

Vì \(\left(2x-\frac{5}{2}\right)^2\ge0\Rightarrow2B=\left(2x-\frac{5}{2}\right)^2-\frac{41}{4}\ge-\frac{41}{4}\Rightarrow B\ge-\frac{41}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmin=-41/8 khi x=5/4

c,\(C=x^2+5y^2+2xy-y+3=\left(x^2+2xy+y^2\right)+\left(4y^2-y+\frac{1}{16}\right)+\frac{47}{16}=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\)

Vì\(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(2y-\frac{1}{4}\right)^2\ge0\end{cases}}\Rightarrow\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2\ge0\)

\(\Rightarrow C=\left(x+y\right)^2+\left(2y-\frac{1}{4}\right)^2+\frac{47}{16}\ge\frac{47}{16}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\2y-\frac{1}{4}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{8}\\y=\frac{1}{8}\end{cases}}}\)

Vậy Cmin=47/16 khi x=-1/8,y=1/8

(x-1)(x\(^5\)+x\(^4\)+x\(^3\)+x\(^2\)+x+1)

=x(\(x^5+x^4+x^3+x^2+x+1\))-1(\(x^5+x^4+x^3+x^2+x+1\))

= x.\(x^5+x\cdot x^4+x\cdot x^3+x\cdot x^2+x\cdot x+x\cdot1\)-1.\(x^5-1\cdot x^4-1\cdot x^3-1\cdot x^2-1\cdot x-1\cdot1\)

=\(x^6\)+\(x^5\)\(+x^4\)+\(x^3\)+\(x^2\)+1x -1\(x^5\)-1\(x^4\)-1\(x^3\)-1\(x^2\)-1x -1

=\(x^6\)+(\(x^5\)-1\(x^5\))+(x\(^4\)-1\(x^4\))+(\(x^3\)-1\(x^3\))+(x\(^2\)-1\(x^2\))+(1x-1x)-1

=x\(^6\)-1

thanks