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Có: x,y,z tỉ lệ với 5;4;3
\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\)
\(\Rightarrow x=5k;y=4k;z=3k\)
\(P=\frac{x+2y-3z}{x-2y+3z}\)
\(\Rightarrow P=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}\)
\(\Leftrightarrow P=\frac{4k}{6k}\)
\(\Leftrightarrow P=\frac{2}{3}\)
Vậy \(P=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x; y; z tỉ lệ với 5; 4; 3
\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
\(\Rightarrow\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}\)
\(\Rightarrow\frac{x+2y-3z}{5+8-9}=\frac{x-2y+3z}{5-8+9}=\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
\(\Rightarrow\frac{x+2y-3z}{4}=\frac{x-2y+3z}{6}\)
\(\Rightarrow\frac{x+2y-3z}{x-2y+3z}=\frac{4}{6}=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x: y : z = 5 : 4 : 3
đặt x = 5k => y = 4k; z = 3k
P = \(\frac{x+2y-3z}{x-2y+3z}=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{4}{6}=\frac{2}{3}\)
Vậy P = 2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì x,y,z tỉ lệ với 5,4,3 nên \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)=> \(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}=\frac{x+2y-3z}{5+8-9}=\frac{x+2y-3z}{4}\)
và \(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}=\frac{x-2y+3z}{5-8+9}=\frac{x-2y+3z}{6}\)
Từ 2 cái này suy ra : \(\frac{x+2y-3z}{4}=\frac{x-2y+3z}{6}\)=> P=\(\frac{x+2y-3z}{x-2y+3z}=\frac{4}{6}=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\Rightarrow x=5k;y=4k;z=3k\)
=>\(P=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:x:y:z=5:4:3
=>x/5=y/4=z/3
\(=>\frac{x+2y-3z}{5+2.4-3.3}=\frac{x-2y+3z}{5-2.4+3.3}=>\frac{x+2y-3z}{4}=\frac{x-2y+3z}{6}\)
\(=>\frac{x+2y-3z}{x-2y+3z}=\frac{4}{6}=\frac{2}{3}\)
vậy P=2/3
tick nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\)
\(\Rightarrow x=5k\); \(y=4k\); \(z=3k\)
\(\Rightarrow D=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2\left(4k\right)-3\left(3k\right)}{5k-2\left(4k\right)+3\left(3k\right)}\)
\(D=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{2}{3}\)
VẬY, \(D=\frac{2}{3}\)