Ta có:

\(x\left(x+y+z\right)=\frac{15}{2}\)

\(y\left(x+y+z\right)=\frac{-5}{2}\)

\(z\left(x+y+z\right)=20\)

=>\(x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=\frac{15}{2}+\frac{-5}{2}+20\)

                                               \(\left(x+y+z\right)\left(x+y+z\right)=\frac{15-5}{2}+20\)

                                                                     \(\left(x+y+z\right)^2=\frac{10}{2}+20\)

                                                                     \(\left(x+y+z\right)^2=5+20\)

                                                                     \(\left(x+y+z\right)^2=25\)

=>x+y+z=5 hoặc x+y+x=-5

Với x+y+z=5

=>\(x.5=\frac{15}{2}\)=>\(x=\frac{15}{2}.\frac{1}{5}=\frac{3}{2}\)

   \(y.5=\frac{-5}{2}\)=>\(y=\frac{-5}{2}.\frac{1}{5}=\frac{-1}{2}\)

   \(z.5=20\)=>\(z=\frac{20}{5}=4\)

Với x+y+z=-5

=>\(x.\left(-5\right)=\frac{15}{2}\)=>\(x=\frac{15}{2}.\frac{-1}{5}=\frac{-3}{2}\)

   \(y.\left(-5\right)=\frac{-5}{2}\)=>\(y=\frac{-5}{2}.\frac{-1}{5}=\frac{1}{2}\)

   \(z.\left(-5\right)=20\)=>\(z=\frac{20}{-5}=-4\)

Vậy \(x=\frac{3}{2},y=-\frac{1}{2},z=4\);  \(x=-\frac{3}{2},y=\frac{1}{2},z=-4\)

Ta có:

\(x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=\frac{15}{2}+\left(-\frac{5}{2}\right)+20\)(Cộng vế với vế)

\(\Leftrightarrow\left(x+y+z\right)\left(x+y+z\right)=\frac{50}{2}=25\)

\(\Rightarrow\left(x+y+z\right)^2=25\Leftrightarrow x+y+z=\sqrt{25}=5\)

\(\Rightarrow\hept{\begin{cases}x.5=\frac{15}{2}\Rightarrow x=\frac{3}{2}\\y.5=-\frac{5}{2}\Rightarrow y=-\frac{1}{2}\\z.5=20\Rightarrow z=4\end{cases}}\)

Vậy \(x=\frac{3}{2};y=-\frac{1}{2};z=4\).

Theo tính chất dãy tỉ số bằng nhau :

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

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

( Vì x + y + z \(\ne\)0 ) Do đó, x +y + z = 0,5

Thay kết quả này vào đầu đề bài ta được :

\(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}=2\)

tức là

\(\frac{1,5-x}{x}=\frac{2,5-y}{y}=\frac{-2,5-z}{z}=2\)

Vậy \(x=\frac{1}{2},y=\frac{5}{6},z=\frac{-5}{6}\)

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

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

           \(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)

    Tương tự:   \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)

                \(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)

Suy ra: \(A+\left(x+y+z\right)\)

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

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

  \(=2.\left(x+y+z\right)\)

Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)

Mình có sai chỗ nào không nhỉ?