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{ 1 x + 1 y + z = 1 2 1 y + 1 z + x = 1 3 1 z + 1 x + y = 1 4 ⇔ { x + y + z x ( y + z ) = 1 2 x + y + z y ( z + x ) = 1 3 x + y + z z ( x + y ) = 1 4 ⇔ { 2 ( x + y + z ) x y z = x ( y + z ) x y z = 1 y + 1 z 3 ( x + y + z ) x y z = y ( z + x ) x y z = 1 z + 1 x 4 ( x + y + z ) x y z = z ( x + y ) x y z = 1 x + 1 y ( 1 ) \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}\\\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x+y+z}{x(y+z)}=\frac{1}{2}\\\frac{x+y+z}{y(z+x)}=\frac{1}{3} \\\frac{x+y+z}{z(x+y)} =\frac{1}{4} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{2(x+y+z)}{xyz}=\frac{x(y+z)}{xyz}=\frac{1}{y}+\frac{1}{z}\\\frac{3(x+y+z)}{xyz}=\frac{y(z+x)}{xyz}=\frac{1}{z}+\frac{1}{x} \\\frac{4(x+y+z)}{xyz}=\frac{z(x+y)}{xyz}=\frac{1}{x}+\frac{1}{y} \end{matrix}\right.(1) ⎩ ⎪ ⎨ ⎪ ⎧ x 1 + y + z 1 = 2 1 y 1 + z + x 1 = 3 1 z 1 + x + y 1 = 4 1 ⇔ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x ( y + z ) x + y + z = 2 1 y ( z + x ) x + y + z = 3 1 z ( x + y ) x + y + z = 4 1 ⇔ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x y z 2 ( x + y + z ) = x y z x ( y + z ) = y 1 + z 1 x y z 3 ( x + y + z ) = x y z y ( z + x ) = z 1 + x 1 x y z 4 ( x + y + z ) = x y z z ( x + y ) = x 1 + y 1 ( 1 )
Đặt x + y + z x y z = 1 a ( ∗ ) \frac{x+y+z}{xyz}=\frac{1}{a}(*) x y z x + y + z = a 1 ( ∗ )
Thay vào ( 1 ) (1) ( 1 ) ta được { 2 a = 1 y + 1 z 3 a = 1 z + 1 x 4 a = 1 x + 1 y ⇔ { 1 x = 5 2 a 1 y = 3 2 a 1 z = a 2 ⇔ { x = 2 a 5 y = 2 a 3 z = 2 a ( ∗ ∗ ) \left\{\begin{matrix} \frac{2}{a}=\frac{1}{y}+\frac{1}{z}\\ \frac{3}{a}=\frac{1}{z}+\frac{1}{x} \\\frac{4}{a}=\frac{1}{x}+\frac{1}{y} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}=\frac{5}{2a}\\\frac{1}{y}= \frac{3}{2a} \\\frac{1}{z}=\frac{a}{2} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{2a}{5}\\y=\frac{2a}{3} \\z=\frac{2}{a} \end{matrix}\right.(**) ⎩ ⎪ ⎨ ⎪ ⎧ a 2 = y 1 + z 1 a 3 = z 1 + x 1 a 4 = x 1 + y 1 ⇔ ⎩ ⎪ ⎨ ⎪ ⎧ x 1 = 2 a 5 y 1 = 2 a 3 z 1 = 2 a ⇔ ⎩ ⎪ ⎨ ⎪ ⎧ x = 5 2 a y = 3 2 a z = a 2 ( ∗ ∗ )
Thay trở lại ( ∗ ) (*) ( ∗ ) , tìm được a a a rồi tìm được x , y , z x,y,z x , y , z