Question

[Phạm Hữu Dũng]

Tìm x; y; z. Biết :

a.  4/ x+1=2/y-2=3/z+2 và  x+y+z =17 

Answer 1

[✎✰ ๖ۣۜLαɗσηηα ༣✰✍]

 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!!

Answer 2

[Gukmin]

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