Question

[huy bình]

tìm x;y;z

xy+z+1xy+z+1 = yx+z+1yx+z+1 =zx+y−2zx+y−2 =x+y+z

Answer 1

[Lightning Farron]

Áp dụng tc dãy tỉ số bằng nhau ta có:

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

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

Lại có:

• \(\frac{z}{x+y-2}=\frac{1}{2}\Rightarrow2z=x+y-2\Rightarrow2z+2=x+y\)

\(\Rightarrow x+y+z=\frac{1}{2}\)\(\Leftrightarrow2z+2+z=\frac{1}{2}\Leftrightarrow3z=\frac{1}{2}-2=-\frac{3}{2}\Rightarrow z=-\frac{1}{2}\)

• \(\frac{y}{x+z+1}=\frac{1}{2}\Rightarrow2y=x+z+1\Rightarrow2y-1=x+z\)

\(\Rightarrow x+y+z=\frac{1}{2}\Leftrightarrow y+2y-1=\frac{1}{2}\Leftrightarrow3y=\frac{3}{2}\Rightarrow y=\frac{1}{2}\)

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

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

Answer 2

[Lightning Farron]

đề đúnh

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