Question

Tìm x,y,z biết

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

làm đúng 3 tick :))

Answer 1

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

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

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

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

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

\(\Rightarrow2x=y+z-2\)

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

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

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

\(\Rightarrow2y=x+z+1\)

\(\Rightarrow3y=x+y+z+1=\frac{1}{2}+1=\frac{3}{2}\)

\(\Rightarrow y=\frac{1}{2}\)

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

Vậy ...

Answer 2

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