Bài 1:
Giải:
Ta có: \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\)
\(5x=7z\Rightarrow\dfrac{x}{7}=\dfrac{z}{5}\Rightarrow\dfrac{x}{21}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{75}=\dfrac{3x-7y+5z}{63-98+75}=\dfrac{30}{40}=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{63}{4}\\y=\dfrac{21}{2}\\z=\dfrac{45}{4}\end{matrix}\right.\)
Vậy...
Bài 2:
Giải:
Ta có: \(3a=2b\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{15}\)
\(4b=5c\Rightarrow\dfrac{b}{5}=\dfrac{c}{4}\Rightarrow\dfrac{b}{15}=\dfrac{c}{12}\)
\(\Rightarrow\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}=\dfrac{-a-b+c}{-10-15+12}=\dfrac{-52}{-13}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
Vậy...
Bài 1:
Ta có: \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) (1)
\(5x=7z\Rightarrow\dfrac{x}{7}=\dfrac{z}{5}\Rightarrow\dfrac{x}{21}=\dfrac{y}{15}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
Ta có: \(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{15}\Rightarrow\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{75}=\dfrac{3x-7y+5z}{63-98+75}=\dfrac{30}{40}=\dfrac{3}{4}\)
Khi đó:
+) \(\dfrac{x}{21}=\dfrac{3}{4}\Rightarrow x=\dfrac{21\cdot3}{4}=\dfrac{63}{4}\)
+) \(\dfrac{y}{14}=\dfrac{3}{4}\Rightarrow y=\dfrac{14\cdot3}{4}=\dfrac{21}{2}\)
+) \(\dfrac{z}{15}=\dfrac{3}{4}\Rightarrow z=\dfrac{15\cdot3}{4}=\dfrac{45}{4}\)
Vậy \(x=\dfrac{63}{4};y=\dfrac{21}{2};z=\dfrac{45}{4}\)
