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:

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

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

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

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

avt ảnh bạn à, vừa handsome vừa học giỏi nx -.-

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                                    

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

=> x+y+z =1/2

+y+z+1=2x  => x+y+z +1 =3x => 3x =1/2 +1 =3/2 => x =1/2

+x+y+2 =2y  => x+y+z+2 =3y  => 3y = 1/2 +2 = 5/2 => y =5/6

+z =1/2 -x-y =1/2 -1/2 -5/6 =-5/6

lai 1 thag nua dug chieu xin lik e = cah do nua a T_T