\(E=\frac{5}{2x^2+3x+5}=\frac{5}{2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{35}{8}}=\frac{5}{2\left(x+\frac{3}{4}\right)^2+\frac{35}{8}}\le\frac{5}{\frac{35}{8}}=\frac{8}{7}\)

Nên GTLN của E là \(\frac{8}{7}\) đạt được khi x=\(-\frac{3}{4}\)

\(F=\frac{-2}{4x-x^2-5}=\frac{2}{x^2-4x+5}=\frac{2}{x^2-2.2x+4+1}=\frac{2}{\left(x-2\right)^2+1}\le\frac{2}{1}=2\)

Nên GTLN của F là 2 đạt được khi \(x=2\)

GTLN cua F la 2 khi 

x=2 

chuc ban hoc tot

a,\(x^2y^2+y^3+zx^2+yz=\left(x^2y^2+y^3\right)+\left(zx^2+yz\right)\)

\(=y^2\left(x^2+y\right)+z\left(x^2+y\right)\)

\(=\left(y^2+z\right)\left(x^2+y\right)\)

b,\(x^4+2x^3-4x-4=x^4+2x^3+x^2-x^2-4x-4\)

\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

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

c,\(x^3+2x^2y-x-2y=\left(x^3+2x^2y\right)-\left(x+2y\right)\)

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

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

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

Mình có làm cho bạn mấy bài trước rồi nên nhìn vào mấy bài đó để làm tương tự nhé (giờ làm tắt)

+)\(P=2x-x^2=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)

Dấu "=" xảy ra khi x=1

Vậy Pmax=1 khi x=1

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

Dấu "=" xảy ra khi x=1/2

Vậy Qmax=5/4 khi x=1/2

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

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

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

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

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

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

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

=> \(M\le20\)\(\forall x\)

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