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:

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

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

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

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

Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)

tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)

=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)

Dấu "=" xảy ra khi x=y=z=4

Vậy minM=6 khi x=y=z=4

b1: Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

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

=>minP=1 <=> x=y=z=2/3

y+x+z bằng bao nhiêu mới tính ra được chứ?? sai đề à??

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

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

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

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.

Điều kiện \(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)

ADTC dãy tỉ số bằng nhau ta có : 

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

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

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

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

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

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

\(\Leftrightarrow x=\frac{1}{2}\Rightarrow y+z=0\Leftrightarrow y=-z\)

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

\(\Leftrightarrow x+y-3=2z\)

\(\Leftrightarrow y-2z=\frac{5}{2}\)

Do: \(y=-z\Rightarrow-3z=\frac{5}{2}\Leftrightarrow z=-\frac{5}{6}\)

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

Vậy nghiệm tìm đc : \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{5}{6};-\frac{5}{6}\right)\)

M = x+y/z + x+z/y + y+z/x

M = x+y+z/z + x+y+z/y + x+y+z/x - z/z - y/y - x/x

M = (x+y+z).(1/z + 1/y + 1/x) - 1 - 1 - 1

M = 2020.1/202 - 3

M = 10 - 3 = 7

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

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

<=>\(\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{xy}{x+z}+\frac{y^2}{x+z}+\frac{yz}{x+z}+\frac{xz}{x+y}+\frac{yz}{x+y}+\frac{z^2}{x+y}=1\)

<=>\(\frac{x^2}{y+z}+\frac{xy+xz}{y+z}+\frac{y^2}{x+z}+\frac{xy+yz}{x+z}+\frac{z^2}{x+y}+\frac{xz+yz}{x+y}=x+y+z\)

<=>\(\frac{x^2}{y+z}+\frac{x\left(y+z\right)}{y+z}+\frac{y^2}{x+z}+\frac{y\left(x+z\right)}{x+z}+\frac{z^2}{x+y}+\frac{z\left(x+y\right)}{x+y}=x+y+z\)

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

<=>\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)