Ta có: \(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\Leftrightarrow\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}\)
Đặt \(\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}=k\left(k\ne0\right)\)
\(\Rightarrow x=4k-1;y=2k+2;z=3k-2\)
Theo đề ta có:
\(x+y+z=17\)
hay \(4k-1+2k+2+3k-2=17\)
\(9k-1=17\)
\(9k=18\)
\(k=\frac{18}{9}=2\)
Do đó:
\(x=4.2-1=8-1=7\)
\(y=2.2+2=4+2=6\)
\(z=3.2-2=6-2=4\)
Vậy \(x=7;y=6;z=4\)
hok tốt!!
Trả lời:
\(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\)\(\left(Đk:x\ne-1;y\ne2;z\ne-2\right)\)
\(\Leftrightarrow\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}\)
Đặt\(\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x+1=4k\\y-2=2k\\z+2=3k\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4k-1\\y=2k+2\\z=3k-2\end{cases}}\)
Mà\(x+y+z=17\)
\(\Rightarrow4k-1+2k+2+3k-2=17\)
\(\Leftrightarrow9k-1=17\)
\(\Leftrightarrow9k=18\)
\(\Leftrightarrow k=2\)
\(\Rightarrow\hept{\begin{cases}x=2.4-1=7\\y=2.2+2=6\\z=2.3-2=4\end{cases}}\)(Thỏa mãn\(Đk:x\ne-1;y\ne2;z\ne-2\))
Vậy\(\hept{\begin{cases}x=7\\y=6\\z=4\end{cases}}\)
Hok tốt!
Good girl
