ta có \(\frac{2}{\sqrt{x}}-z=\frac{2\sqrt{xyz}}{\sqrt{x}}-z\)\(=2\sqrt{yz}-z\le y+z-z=y\)THEO bđt côsi

Tương tự \(\frac{2}{\sqrt{y}}-x\le z\)và \(\frac{2}{\sqrt{z}}-y\le x\)

\(\Rightarrow A\le xyz=1\)

VẬY MAX A=1 TẠI x=y=z=1

quang phan duy Sol hay đấy =) hay hơn cách tôi rồi

a) ĐK : \(x\ne1;x\ne2;x\ne3\)

\(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)

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

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

\(\Leftrightarrow K=\frac{2x^2}{x^4+x^2+1}\)

a, \(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-2\right)}{x^4+x^2+1}\) 

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

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

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

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

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

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

\(=\frac{2x^2}{x^4+x^2+1}\)

b)  +) Trường hợp 1 :

Nếu \(x=0\)

\(\Rightarrow K=0\)

+) Trường hợp 2 :

Nếu \(x\ne0\)

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

Vậy để K đạt GTLN  khi x=2/3 \(\Leftrightarrow\)x=-1

NHỚ K MK NHA!!!

a)Áp dụng BĐT (x+y)^2>=4xy>>>(3a+5b)^2>=4.3a.5b>>>144>=60ab>>>ab<=12/5

Dấu=xảy ra khi 3a=5b hay khi a=7,5;b=4.5(không nên dùng Cô-si vì không chắc chắn là số dương).

b)Áp dụng BĐT Cô-si>>>(y+10)^2>=40y(do ở đây y>0 nên có thể dùng Cô-si)>>>A<=y/40y=1/40

Dấu= xảy ra khi y=10.

c)A=(x^2+x+1)/x^2+2x+1=1/2(2x^2+2x+1)/x^2+2x+1>>>A/2=(x^2+2x+1)/(x^2+2x+1)+x^2/(x^2+2x+1))>=1+0=1

Dấu= xảy ra khi x=0

A

Áp dụng BĐT cosi ta có 

\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)

\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)

Khi đó 

\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)

MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)

B

Áp dụng BĐT cosi ta có :

\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)

=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)

=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)

Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)

=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)

=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)

\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z