Ta có: \(3x=5y=7z\)
\(\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{7}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{7}}=\dfrac{x+y-z}{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}}=\dfrac{41}{\dfrac{41}{105}}=105\)
+) \(\dfrac{x}{\dfrac{1}{3}}=105\Rightarrow x=\dfrac{1}{3}.105=35\)
+) \(\dfrac{y}{\dfrac{1}{5}}=105\Rightarrow x=\dfrac{1}{5}.105=21\)
+) \(\dfrac{z}{\dfrac{1}{7}}=105\Rightarrow z=\dfrac{1}{7}.105=15\)
Vậy \(x=35;y=21;z=15\)
Ta có 3x=5y=7z suy ra \(\dfrac{x}{\dfrac{1}{3}}\)=\(\dfrac{y}{\dfrac{1}{5}}\)=\(\dfrac{z}{\dfrac{1}{7}}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có:
\(\dfrac{x}{\dfrac{1}{3}}=\) \(\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{7}}\)\(=\dfrac{x}{\dfrac{1}{3}}+\dfrac{y}{\dfrac{1}{5}}+\dfrac{z}{\dfrac{1}{7}}=\dfrac{x+y-z}{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}}=\dfrac{41}{\dfrac{41}{105}}=105\)suy ra : x = 105 . \(\dfrac{1}{3}\)= 35
y = 105 . \(\dfrac{1}{5}\)= 21
z = 105 . \(\dfrac{1}{7}\)=15
Vậy : ...
