Ta có : \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=\(\dfrac{d}{5}\) và 3a + b- 2c + 4d = 105
Đặt \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)=\(\dfrac{d}{5}\) = k
\(\Rightarrow\): a = 2k; b = 3k; c = 4k; d = 5k
Ta có : 3(2k) + 3k -2(4k) + 4(5k) = 105
6k +3k - 8k + 20k = 105
21k = 105
\(\Rightarrow\) k = 5
+) a = 2.5 = 10
+) b = 3.5 = 15
+) c = 4.5 = 20
+) d = 5.5 = 25
Ta có: \(a:b:c:d=2:3:4:5\) nên \(\Rightarrow a,b,c,d\) lần lượt tỉ lệ với các số \(2,3,4,5\) .
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{c}{5}\)\(\Rightarrow\dfrac{3a}{6}=\dfrac{b}{3}=\dfrac{2c}{8}=\dfrac{4d}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{3a}{6}=\dfrac{b}{3}=\dfrac{2c}{8}=\dfrac{4c}{20}\) \(=\dfrac{3a+b-2c+4d}{6+3-8+20}\)\(=\dfrac{105}{21}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\\\dfrac{b}{3}=5\\\dfrac{c}{4}=5\\\dfrac{d}{5}=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=5.2\\b=5.3\\c=5.4\\d=5.5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\\d=25\end{matrix}\right.\)
Vậy \(a=10;b=15;c=20;d=25\)
Ta có 3a + b =105 + 2c
\(\Rightarrow\) 3a + b - 2c + 4d =105
\(\Rightarrow\)\(\dfrac{a}{2}\)= \(\dfrac{b}{3}\)= \(\dfrac{c}{4}\) = \(\dfrac{d}{5}\)
\(\Rightarrow\)\(\dfrac{3a}{6}\)= \(\dfrac{b}{3}\)=\(\dfrac{2c}{8}\) =\(\dfrac{4d}{20}\)= \(\dfrac{3a+b-2c+4d}{6+3-8+20}\)=\(\dfrac{105}{21}\)=\(5\)
\(\Rightarrow\)a= 5x6:3=10
\(\Rightarrow\)b=5x3=15
\(\Rightarrow\)c=5x8:2=20
\(\Rightarrow\)d=5x20:4=25
