Tìm x,y,z biết
2/3x=3/4y;1/5x=3/7z và 3x +4y-9z=254
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Tìm x,y,z biết
2/3x=3/4y;1/5x=3/7z và 3x +4y-9z=254
Giúp em vs ạ
Em đang cần gấp!!!
Xin chân thành cảm ơn ạ!!!
Đáp án:
$\left( {x;y;z} \right) = \left( {\dfrac{{5715}}{{53}};\dfrac{{5080}}{{53}};\dfrac{{2667}}{{53}}} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\dfrac{2}{3}x = \dfrac{3}{4}y \Rightarrow \dfrac{2}{9}x = \dfrac{1}{4}y \Rightarrow \dfrac{x}{9} = \dfrac{y}{8} \Rightarrow \dfrac{x}{{45}} = \dfrac{y}{{40}}\\
\dfrac{1}{5}x = \dfrac{3}{7}z \Rightarrow \dfrac{1}{{15}}x = \dfrac{1}{7}z \Rightarrow \dfrac{x}{{15}} = \dfrac{z}{7} \Rightarrow \dfrac{x}{{45}} = \dfrac{z}{{21}}\\
\Rightarrow \dfrac{x}{{45}} = \dfrac{y}{{40}} = \dfrac{z}{{21}}\\
\Rightarrow \dfrac{x}{{45}} = \dfrac{y}{{40}} = \dfrac{z}{{21}} = \dfrac{{3x + 4y – 9z}}{{3.45 + 4.40 – 9.21}} = \dfrac{{254}}{{106}} = \dfrac{{127}}{{53}}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{{45}} = \dfrac{{127}}{{53}}\\
\dfrac{y}{{40}} = \dfrac{{127}}{{53}}\\
\dfrac{z}{{21}} = \dfrac{{127}}{{53}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{5715}}{{53}}\\
y = \dfrac{{5080}}{{53}}\\
z = \dfrac{{2667}}{{53}}
\end{array} \right.\\
\Rightarrow \left( {x;y;z} \right) = \left( {\dfrac{{5715}}{{53}};\dfrac{{5080}}{{53}};\dfrac{{2667}}{{53}}} \right)
\end{array}$
Vậy $\left( {x;y;z} \right) = \left( {\dfrac{{5715}}{{53}};\dfrac{{5080}}{{53}};\dfrac{{2667}}{{53}}} \right)$
`+) 2/3 x=3/4 y`
`=> 2/9 x=1/4 y`
`=> x/9=y/8`
`=> x/45=y/40`
`+) 1/5 x=3/7 z`
`=> 1/5 x =1/7z`
`=> x/15=z/7`
`=> x/45=z/21`
`=> x/45=y/40=z/21 =(2x+4y-9z)/(3.45+4.40-9.21)`
`=254/105`
`=127/53 `
`+) x/45=127/53`
`=> x=5715/53`
`+) y/40=127/54`
`=> y=5080/53`
`+) z/21=127/53`
`=> z=2667/53`