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Vì x,y,z tỉ lệ với 5,4,3 nên ta có \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2y}{8}=\dfrac{3z}{9}=\dfrac{x+2y-3z}{5+8-9}=\dfrac{x+2y-3z}{4}\)
Và \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2y}{8}=\dfrac{3z}{9}=\dfrac{x-2y+3z}{5-8+9}=\dfrac{x-2y+3z}{6}\)
Do đó \(\dfrac{x+2y-3z}{4}=\dfrac{x-2y+3z}{6}\)
=> \(\dfrac{x+2y-3z}{x-2y+3z}=\dfrac{4}{6}=\dfrac{2}{3}\)
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 : \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
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, \(D=\frac{2}{3}\)
![](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)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-8y}{16}=0\\\dfrac{2z-4x}{3}=0\\\dfrac{4y-3z}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x-8y=0\\2x-4z=0\\4y-3z=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x,y,z tỉ lệ với 5,4,3
\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
\(\Rightarrow\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}\)
\(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}=\frac{x+2y-3z}{5+8-9}=\frac{x+2y-3z}{4}\)( 1 )
\(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}=\frac{x-2y+3z}{5-8+9}=\frac{x-2y+3z}{6}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x+2y-3z}{4}=\frac{x-2y+3z}{6}\)
\(\Rightarrow\frac{x+2y-3z}{x-2y+3z}=\frac{4}{6}=\frac{2}{3}\)
Vì x,y,z tỉ lệ với 5;4;3, ta có: \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Áp dụng tính chất Dãy tỉ số bằng nhau, ta có:
\(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{2y}{8}=\frac{3z}{9}=\frac{x+2y-3z}{5+8-9}=\frac{x+2y-3z}{4}\)
Và \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{2y}{8}=\frac{3z}{9}=\frac{x-2y+3z}{5-8+9}=\frac{x-2y+3z}{6}\)
Do đó: \(\frac{x+2y-3z}{x-2y+3z}=\frac{x-2y+3z}{6}\)
\(\Rightarrow\frac{x+2y-3x}{x-2y+3x}=\frac{4}{6}=\frac{2}{3}\)
Vậy: \(P=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\\z=3k\end{cases}}\)
Khi đó P = \(\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{2}{3}\)
Theo bài ra, ta có :
x:y:z=5:4:3 ⇒x/5=y/4=z/5⇒
Đặt x/5=y/4=z/3=kx5=y4=z3=k ⇒x=5k
y=4k
z=3k⇒x=5ky=4kz=3k
⇒P=x+2y−3z/x−2y+3z=5k+8k−9k/5k−8k+9k=4k/6k=23
Vậy P=23
Theo bài ra, ta có :
\(x:y:z=5:4:3\) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=3k\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{x+2y-3z}{x-2y+3z}=\dfrac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\dfrac{5k+8k-9k}{5k-8k+9k}=\dfrac{4k}{6k}=\dfrac{2}{3}\)
Vậy \(P=\dfrac{2}{3}\)