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Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{5a-4b}{6}=\frac{6b-5c}{4}=\frac{4c-6a}{5}\)
\(\Rightarrow \frac{6(5a-4b)}{36}=\frac{4(6b-5c)}{16}=\frac{5(4c-6a)}{25}=\frac{6(5a-4b)+4(6b-5c)+5(4c-6a)}{36+16+25}=\frac{0}{36+16+25}=0\)
\(\Rightarrow \left\{\begin{matrix} 5a-4b=0\\ 6b-5c=0\\ 4c-6a=0\end{matrix}\right.\Rightarrow \frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)
Áp dụng TCDTSBN:
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{45}{15}=3\)
\(\Rightarrow \left\{\begin{matrix} a=4.3=12\\ b=5.3=15\\ c=5.3=18\end{matrix}\right.\)
\(\frac{\left(5a-4b\right)6}{36}=\frac{\left(6a-4c\right)5}{25}=\frac{\left(6b-5c\right)4}{16}=\frac{\left(5a-4b\right)6-\left(6a-4c\right)5+\left(6b-5c\right)4}{36-25+16}=\frac{0}{27}\)
\(\Rightarrow5a=4b\Leftrightarrow\frac{a}{4}=\frac{b}{5}\)
\(\Rightarrow6a=4c\Leftrightarrow\frac{a}{4}=\frac{c}{6}\)
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)
\(\hept{\begin{cases}2a+3b+2c=5\\5a+4b+c=55\\a+b-4c=24\end{cases}}\Leftrightarrow8a+8b-c=5+55+24\)
\(\Leftrightarrow8a+8b-c=84\)
\(\Leftrightarrow8\left(a+b\right)-c=84\)
\(\Leftrightarrow8\left(a+b\right)=84+c\)
\(\Rightarrow a+b+c=84\)
\(\Rightarrow TBC\left(a,b,c\right)=\frac{84}{3}=28\)
bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)
áp dụng dảy tỉ số bằng nhau
ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)
\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)
\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)
vậy \(a=-3;b=-11;c=-7\)
b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)
áp dụng dảy tỉ số bằng nhau
ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)
\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
vậy \(a=40;b=60;c=48\)
Theo đầu bài ta có:
\(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)
\(\Rightarrow\frac{5c-25}{30}=\frac{3a-3}{6}=\frac{4b+12}{16}\)
\(=\frac{\left(5c-25\right)-\left(3a-3\right)-\left(4b+12\right)}{30-6-16}\)
\(=\frac{\left(5c-3a-4b\right)-\left(25-3+12\right)}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}a=2\cdot2+1=5\\b=2\cdot4-3=5\\c=2\cdot6+5=17\end{cases}}\)
\(\left\{{}\begin{matrix}2a+3b+2c=5\\5a+4b+c=55\\a+b-4c=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10a+15b+10c=25\\10a+8b+2c=110\\10a+10b-40c=240\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10a+15b+15c=25\\7b+8c=-85\\5b+50c=-215\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+3b+2c=5\\b=-\dfrac{253}{31}\\c=-\dfrac{108}{31}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{565}{31}\\b=-\dfrac{253}{31}\\c=-\dfrac{108}{31}\end{matrix}\right.\)
Trung bình cộng của ba số là:
\(\left(\dfrac{565}{31}-\dfrac{253}{31}-\dfrac{108}{31}\right):3=\dfrac{68}{31}\)