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Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\Rightarrow x=5k;y=4k;z=3k\)
=>\(P=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{2}{3}\)
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 = \(\frac{2}{3}\)
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
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2y}{8}=\dfrac{3z}{9}=\dfrac{x+2y-3z}{5+8-9}=\dfrac{x-2y+3z}{5-8+9}\\ \Rightarrow A=\dfrac{x+2y-3z}{x-2y+3z}=\dfrac{4}{6}=\dfrac{2}{3}\)
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 \(P=\frac{2}{3}\)
D = \(\frac{2}{3}\) .
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, \(D=\frac{2}{3}\)