\(\sqrt{x^2}=\left|-9\right|\)

\(\Leftrightarrow x=9\)

x=-9 nữa

\(2\sqrt{x}-x-3=-x+2\sqrt{x}-1-2=-\left(\sqrt{x}-1\right)^2-2\)

Ta có: \(\left(\sqrt{x}-1\right)^2\ge0\Rightarrow-\left(\sqrt{x}-1\right)^2\le0\Rightarrow-\left(\sqrt{x}-1\right)^2-2\le-2\)

=> \(A=\frac{1}{2\sqrt{x}-x-3}\ge-\frac{1}{2}\)

Dấu"=" xảy ra <=> \(\sqrt{x}-1=0\)<=> x = 1

Vậy max A = -1/2 đạt tại x = 1.

Hướng dẫn: 

Gọi F là giao điểm của d và AB

\(\Delta\)BFE ~ \(\Delta\)DBA ( g - g - g) 

=> \(\frac{BF}{DB}=\frac{BE}{DA}\)=> BF . DA = DB . BE  (1) 

Ta có : BD // CF => \(\frac{AB}{BF}=\frac{AD}{DC}\)=> AB . DC = AD . BF  (2)

Từ (1) ; (2) => DB . BE = AB . DC => \(\frac{BD}{AB}=\frac{DC}{BE}\)(3)

Có:  CF // BD và BE vuông CF => BE vuông DB => ^DBE = 90\(^o\)

=> ^EBF  + ^DBA = 90\(^o\)

mà ^DBA + ^ADB = 90\(^o\)

=> ^EBF = ^ADB 

=> ^CDB = ^EBA ( 4 )

3, 4 => \(\Delta\)BAE ~ \(\Delta\)DBC ( c.g.c)

 \(A=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

\(A=\sqrt{2-2\sqrt{2}.1+1}-\sqrt{4+2.2\sqrt{2}+2}\)

\(A=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

\(A=\left|\sqrt{2-1}\right|-\left|2+\sqrt{2}\right|\)

\(A=\sqrt{2}-1-2-\sqrt{2}\)|

\(A=-3\)

\(A=\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

\(=\sqrt{2}-1-\left(2+\sqrt{2}\right)\)

\(=\sqrt{2}-1-2-\sqrt{2}\)

\(=-1-2\)

\(=-3\)

A B C D E H M

Kẻ HM vuông góc BC ( M thuộc BC )

\(\Delta BHM~\Delta BCD\left(g.g\right)\) \(\Rightarrow\frac{BH}{BC}=\frac{BM}{BD}\Rightarrow BH.BD=BC.BM\)  ( 1 )

\(\Delta CHM~\Delta CBE\left(g.g\right)\Rightarrow\frac{CH}{BC}=\frac{CM}{CE}\Rightarrow CH.CE=BC.CM\)   ( 2 )

Từ ( 1  ) và ( 2 ) \(\Rightarrow BH.BD+CH.CE=BC\left(BM+CM\right)=BC^2\)

Cô-si Engel :

\(P=\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+6}=\frac{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{a+b+c+6}\)

\(\ge\frac{a+b+c+2.3\sqrt[3]{\sqrt{ab}.\sqrt{bc}.\sqrt{ac}}}{a+b+c+6}=\frac{a+b+c+6\sqrt[3]{abc}}{a+b+c+6}=\frac{a+b+c+6}{a+b+c+6}=1\)

Nguyễn Linh Chi Thanks cô,e đổi biến lộn ạ:(

Đặt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\)

Ta có:

\(P=\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\)

\(=\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\)

\(=\frac{1}{1+\frac{2y}{x}}+\frac{1}{1+\frac{2z}{y}}+\frac{1}{1+\frac{2x}{z}}\)

\(=\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\)

\(=\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2zx}\)

\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Theo Bunhiacopski ta luôn có:

\(\left(x-y\right)^2=\left[1\cdot x+\left(-\frac{1}{2}\right)\cdot2y\right]^2\le\left(1^2+\frac{1}{4}\right)\left(x^2+4y^2\right)=\frac{5}{2}\)

\(\Rightarrow\left|x-y\right|\le\frac{\sqrt{5}}{2}\left(đpcm\right)\)