Bài 1:
b. Ta có:
\(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}\)
\(4y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{10}=\dfrac{z}{8}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{4x-3y+5z}{60-30+40}=\dfrac{7}{70}=\dfrac{1}{10}\)
\(\Rightarrow x=\dfrac{1}{10}\cdot15=\dfrac{3}{2};y=\dfrac{1}{10}\cdot10=1;z=\dfrac{1}{10}\cdot8=\dfrac{4}{5}\)
Vậy \(x=\dfrac{3}{2};y=1;z=\dfrac{4}{5}\)
c. Đặt \(\dfrac{2x}{3}=\dfrac{2y}{4}=\dfrac{4z}{5}=a\)
\(\Rightarrow2x=3a;2y=4a;4z=5a\)
\(\Rightarrow x=\dfrac{3a}{2};y=2a;z=\dfrac{5a}{4}\)
\(\Rightarrow x+y+z=\dfrac{3a}{2}+2a+\dfrac{5a}{4}=49\)
\(\Rightarrow\dfrac{6a}{4}+\dfrac{8a}{4}+\dfrac{5a}{4}=49\)
\(\Rightarrow\dfrac{19a}{4}=49\Rightarrow19a=49\cdot4=196\)
\(\Rightarrow a=\dfrac{196}{19}\)
\(\Rightarrow x=\dfrac{\dfrac{196}{19}\cdot3}{2}=\dfrac{294}{19};y=\dfrac{196}{19}\cdot2=\dfrac{392}{19};z=\dfrac{\dfrac{196}{19}\cdot5}{4}=\dfrac{245}{19}\)
Vậy \(x=\dfrac{294}{19};y=\dfrac{392}{19};z=\dfrac{245}{19}\)
