Question

\(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\)   và x+y+z=12    Tính x,y,z

Answer 1

\(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\Leftrightarrow\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}\)

Đặt \(\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}=k\)\(\Rightarrow\hept{\begin{cases}x=4k-1\\y=2k+2\\z=3k-2\end{cases}}\)

Theo đề bài: x + y + z = 12 <=> \(4k-1+2k+2+3k-2=12\)

<=>\(9k-1=12\Leftrightarrow9k=13\Leftrightarrow k=\frac{13}{9}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{13}{9}.4-1=\frac{43}{9}\\y=\frac{13}{9}.2+2=\frac{44}{9}\\z=\frac{13}{9}.3-2=\frac{7}{3}\end{cases}}\)

Vậy .....................

Answer 2

I don't know