Question

cho x/10 = y/15 = z/21 

tính 3x-y+2z / 2x+5y-7z

 

Answer 1

Đặt \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=k\Rightarrow\begin{cases}x=10k\\y=15k\\z=21k\end{cases}\)

Ta có:

\(\frac{3x-y+2z}{2x+5y-7z}=\frac{3.10k-15k+2.21k}{2.10k+5.15k-7.21k}=\frac{30k-15k+42k}{20k+75k-147k}=\frac{57k}{-52k}=\frac{-57}{52}\)

Vậy \(\frac{3x-y+2z}{2x+5y-7z}=\frac{-57}{52}\)

Answer 2

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{3x}{30}=\frac{2z}{42}=\frac{3x-y+2z}{30-15+42}=\frac{3x-y+2z}{57}\)

=\(\frac{2x}{20}=\frac{5y}{75}=\frac{7z}{147}=\frac{2x+5y-7z}{20+75-147}=\frac{2x+5y-7z}{-52}\)

->\(\frac{3x-y+2z}{2x+5y-7z}=\frac{57}{-52}\)