Ta có :
\(3a+b-2c+4d=105\)
\(a:b:c:d=2:3:4:5\)
\(\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{d}{5}\)
\(\Leftrightarrow\dfrac{3a}{6}=\dfrac{b}{3}=\dfrac{2c}{8}=\dfrac{4d}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{3a}{6}=\dfrac{b}{3}=\dfrac{2c}{8}=\dfrac{4d}{20}=\dfrac{3a+b-2c+4d}{6+3-8+20}=\dfrac{105}{21}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\Leftrightarrow a=10\\\dfrac{b}{3}=5\Leftrightarrow b=15\\\dfrac{c}{4}=5\Leftrightarrow c=20\\\dfrac{d}{5}=5\Leftrightarrow d=25\end{matrix}\right.\)
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