Tìm \(MAX\)

Ta có: \(\frac{2x+1}{x^2+2}=\frac{x^2+2-x^2+2x-1}{x^2+2}\)

\(=1-\frac{\left(x-1\right)^2}{x^2+2}\le1\)

Dấu "=" xảy ra khi \(\Leftrightarrow-\frac{\left(x-1\right)^2}{x^2+2}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy GTLN của biểu thức là \(1\) tại \(x=1\)

Tìm \(MIN\)

Ta có: \(1-\frac{\left(x-1\right)^2}{x^2+2}=-\frac{1}{2}+\frac{3}{2}-\frac{\left(x-1\right)^2}{x^2+2}\)

\(=-\frac{1}{2}+\frac{3x^2+6-2x^2+4x-2}{2\left(x^2+2\right)}\)

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

Dấu "=" xảy ra \(\Leftrightarrow\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\) 

Vậy GTNN của biểu thức là \(-\frac{1}{2}\) tại \(x=-2\)

a, \(A=x^2-6x+11\)

\(=x^2-2.3.x+9+2\)

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

Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(MinA=3\Leftrightarrow x=3\)

b, \(B=2x^2+10x-1\)

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

\(=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{21}{4}\)

\(=2\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\)

Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MinB=-\frac{21}{4}\Leftrightarrow x=-\frac{5}{2}\)

c, \(C=5x-x^2\)

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

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

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

Ta có: \(-\left(x+\frac{5}{2}\right)^2\le0\Leftrightarrow-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MaxB=\frac{25}{4}\Leftrightarrow x=-\frac{5}{2}\)

Lam giup minh voi

\(4B=4x^2+4xy+4y^2-8x-12y+8076\)

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

\(+\left(2x\right)^2-8x+8076\)

= \(\left(2y-3+x\right)^2+3x^2-2x+8076\)

đến đây thì dễ rồi

1,2 kiểu gì ẹ

3,

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge2\)

=> \(\frac{1}{x+1}\ge\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)

Làm tương tự rồi nhân lại ta được \(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

=> \(xyz\le\frac{1}{8}\).Dấu bằng khi x=y=z=1/2

4.

Ta đi CM: \(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{a^2}{a^2+b^2+c^2}\) <=> \(a^4+a\left(b+c\right)^3\le\left(a^2+b^2+c^2\right)^2\)

<=> \(a\left(b+c\right)^3\le2a^2\left(b^2+c^2\right)+\left(b^2+c^2\right)^2\)

Áp dụng BDT COSI thì

\(2a^2\left(b^2+c^2\right)+\left(b^2+c^2\right)^2\ge a^2\left(b+c\right)^2+\frac{\left(b+c\right)^2}{4}\ge a\left(b+c\right)^3\)

Do đó có dpcm

Làm tương tự rồi cộng lại ta đc bdt ban đầu

Dấu bằng xảy ra khi a=b=c

con 2 chưa cho dương nhờ

giúp đê mọi người....