Ta có: 2x=3y=>x/3=y/2 => 15x/15=y/10
4y=5z => y/5=z/4 => y/10 = z/8
Đặt x/15 =y/10 = z/8 = k
=> x = 15k , y = 10k , z = 8k
Ta có : 4(15k) - 3(10k) + 5(8k) = 7
=> 60k - 30k + 40k = 7
=> 70k = 7 => k = 10
Vậy :
x/15 = 10 => x = 10 . 15 = 150
y/10 = 10 => y = 10 . 10 =100
z/4 = 10 => z= 10 . 4 = 40
Có:
\(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)
\(4y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{4}\)
Vì BCNN (2;5) = 10
\(\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{3.5}=\frac{y}{2.5}=\frac{x}{15}=\frac{y}{10}\)
\(\frac{y}{5}=\frac{z}{4}\Leftrightarrow\frac{y}{5.2}=\frac{z}{4.2}=\frac{y}{10}=\frac{z}{8}\)
⇒\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{4x}{60}=\frac{3y}{30}=\frac{5z}{40}=\frac{4x-3y+5z}{60-30+40}=\frac{7}{70}=\frac{1}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.\frac{1}{10}=\frac{3}{2}\\y=10.\frac{1}{10}=1\\z=8.\frac{1}{10}=\frac{4}{5}\end{matrix}\right.\)
