\(e,\) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3};\dfrac{b}{5}=\dfrac{c}{7}\)
\(\Rightarrow\dfrac{a}{10}=\dfrac{b}{15};\dfrac{b}{15}=\dfrac{c}{21}\)
\(\Rightarrow\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{21}\) và \(a+b+c=92\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{21}=\dfrac{a+b+c}{10+15+21}=\dfrac{92}{46}=2\)
Do đó:
\(\dfrac{a}{10}=2\) nên \(a=2.10=20\)
\(\dfrac{b}{15}=2\) nên \(b=2.15=30\)
\(\dfrac{c}{21}=2\) nên \(c=2.21=42\)
Vậy \(a=20;b=30;c=42\)
e: \(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}\\\dfrac{b}{5}=\dfrac{c}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{10}=\dfrac{b}{15}\\\dfrac{b}{15}=\dfrac{c}{21}\end{matrix}\right.\)
=>\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{21}\)
mà a+b+c=92
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{21}=\dfrac{a+b+c}{10+15+21}=\dfrac{92}{46}=2\)
=>\(\left\{{}\begin{matrix}a=2\cdot10=20\\b=2\cdot15=30\\c=2\cdot21=42\end{matrix}\right.\)
d: Đặt \(\dfrac{a-1}{2}=\dfrac{b+2}{3}=\dfrac{c-3}{4}=k\)
=>\(\left\{{}\begin{matrix}a-1=2k\\b+2=3k\\c-3=4k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2k+1\\b=3k-2\\c=4k+3\end{matrix}\right.\)
a+b-2c=-18
=>\(2k+1+3k-2-2\left(4k+3\right)=-18\)
=>\(5k-1-8k-6=-18\)
=>-3k=-18+7=-11
=>\(k=\dfrac{11}{3}\)
=>\(\left\{{}\begin{matrix}a=2\cdot\dfrac{11}{3}+1=\dfrac{22}{3}+1=\dfrac{25}{3}\\b=3\cdot\dfrac{11}{3}-2=11-2=9\\c=4\cdot\dfrac{11}{3}+3=\dfrac{44}{3}+3=\dfrac{53}{3}\end{matrix}\right.\)
