\(P=\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}\)

\(\ge3\sqrt[3]{\frac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\)

Mà theo BĐT AM - GM ta có tiếp:

\(xyz\le\left(\frac{x+y+z}{3}\right)^3=1\)

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\le\left(\frac{x+y+z+3}{3}\right)^3=8\)

\(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra tại x=y=z=1

Vậy..................

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

\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)

a)  \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)

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

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

Thay vào lần lượt ta có:

\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)

\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)

\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)

Cauchy ngược dấu:v

\(A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)\)

\(=x+y+z-\frac{xy+yz+zx}{2}\ge3-\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Đẳng thức xảy ra khi x = y = z = 1.

P/s: Ko chắc~

Bài 1: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

=>\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

=>x=27;z=36;z=60

Bài 2: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\Rightarrow xy=2k.5k=10k^2=40\Rightarrow k^2=4\Rightarrow\hept{\begin{cases}k=-2\\k=2\end{cases}}\)

+)k=-2 => x=-4;y=-5

+)k=2 => x=4;y=5

Vậy x=-4;y=-5 hoặc x=4;y=5

\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)=\(\frac{1}{x+y+z}\)

\(\frac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}\)=\(\frac{1}{x+y+z}\)

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

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

Từ(1) => \(\frac{1}{x+y+z}\)=2 => x+y+z=0,5=>x+z=0,5-y(2)

Từ(1)=> x+y+1=2x(3)

             x+z+2=2y(4)

            z+y-3=2z(5)

Thay(2) vào (4) ta được: 0,5-y+2=2y

                              =>    2,5=3y

                             => y=\(\frac{5}{6}\)

Thay y=\(\frac{5}{6}\)vào(3) ta được:x+\(\frac{5}{6}\)+1=2x

                                            \(\frac{11}{6}\)=x

Thay x=\(\frac{11}{6}\); y=\(\frac{5}{6}\)vào x+y+z=0,5 ta đươc:

\(\frac{11}{6}\)+\(\frac{5}{6}\)+z=0,5

z=\(\frac{-13}{6}\)

      Vậy ............

chúc bn học tốt.

k cho mik nha                                    

Bạn tra trên mạng là có ngay.