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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
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 tỉ lệ với 5,4,3
=> \(\frac{x}{5}\)=\(\frac{y}{4}\)=\(\frac{z}{3}\)
=> x=5.k , y=4.k , z=3.k
=> y =\(\frac{x+2y-3z}{x-2y+3z}\)= \(\frac{5k+2.\left(4k\right)-3.\left(3k\right)}{5k-2.\left(4k\right)+3.\left(3k\right)}\)= \(\frac{5k+8k-9k}{5k-8k+9k}\)= \(\frac{4k}{6k}\)= \(\frac{2}{3}\)
vậy y = \(\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c1:Thay số
Q=\(\frac{5+2.4-3.3}{5-2.4+3.3}\)
O=\(\frac{4}{6}\)=\(\frac{2}{3}\)
\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)
\(4x\left(3y+1\right)=6y\left(2x+8\right)\)
\(12xy+4x=12xy+48y\)
\(4x-48y=0\)
\(4x=48y\)
Ta có:\(\frac{4x}{48y}\)
\(\Leftrightarrow\)\(\frac{x}{y}=\frac{1}{12}\)