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Ta thất:15a=10b=6c
\(\Rightarrow\frac{a}{10}=\frac{b}{15},\frac{b}{6}=\frac{c}{10}\)
\(\Rightarrow\frac{a}{20}=\frac{b}{30}\),\(\frac{b}{30}=\frac{c}{50}\)
Đặt a=29k,b=30k,c=50k
Mà abc=-1920\(\Leftrightarrow k^3=\frac{-8}{125}\)
\(\Leftrightarrow k=\frac{-2}{5}\)
\(\Leftrightarrow\hept{\begin{cases}a=20k=-8\\b=30k=-12\\c=50k=-20\end{cases}}\)
ta có: 15a=10ob=6c
Suy ra a/10=b/15 ; b/6=c/10
Suy ra a/20=b/30 ;b/30=c/50 suy ra a/20=b/30=c/50
Đặt: a/20=b/30=c/50=k
Suy ra a=20k;=30k;c=50k
Mà a,b,c=-1920
20k.30k.50k=-1920
k^3=-8/125 suy ra k=-2/5
suy ra a=20k=-8, b=30k=-12, c=50k=-20
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\(\frac{a}{5}=\frac{b}{8}=\frac{c}{3}\)và a - 2b + c = 34
Ta có:
\(\frac{a}{5}=\frac{b}{8}=\frac{c}{3}=\frac{a}{5}=\frac{2b}{16}=\frac{c}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau,, ta có:
\(\frac{a}{5}=\frac{2b}{16}=\frac{c}{3}=\frac{a-2b+c}{5-16+3}=\frac{34}{-8}=-\frac{17}{4}\)
\(\Rightarrow\frac{a}{5}=-\frac{17}{4}\Rightarrow a=-\frac{17}{4}.5=-\frac{85}{4}\)
\(\Rightarrow\frac{2b}{16}=-\frac{17}{4}\Rightarrow2b=-\frac{17}{4}.16=-68\Rightarrow b=68:2=34\)
\(\Rightarrow\frac{c}{3}=-\frac{17}{4}\Rightarrow c=-\frac{17}{4}.3=-\frac{51}{4}\)
Vậy a = .......
b = ...........
c = .............
\(\frac{a}{5}=\frac{b}{8}=\frac{c}{3}=\frac{a-2b+c}{5-2.8+3}=\frac{34}{-8}\)
=>a=\(\frac{34}{-8}.5=\frac{170}{-8}\)
b=\(\frac{34}{-8}.8=\frac{272}{-8}\)
c=\(\frac{34}{-8}.3=\frac{102}{-8}\)
1 k nhá
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\(15a=10b=6c\Rightarrow c=2,5k;b=1,5k;a=1k\)
\(\Rightarrow a+b+c=2,5k+1,5k+1k=5k=10\Rightarrow k=10:5=2\)
\(\Rightarrow c=2.2,5=5;b=2.1,5=3;a=2.1=2\)
\(Vậy:a=2;b=3;c=5\)
\(15a=10b=6c\Rightarrow1k=1,5k=2,5k\)
\(\Rightarrow a+b+c=1k+1,5k+2.5k=5k=10\)
\(\Rightarrow k=10\div5=2\Rightarrow k=2\)
\(\Rightarrow\hept{\begin{cases}a=2\\b=3\\c=5\end{cases}}\)
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a) Vì BCNN(5;3;8)=120
\(\Rightarrow5a=8b=3c\Leftrightarrow\frac{5a}{120}=\frac{8b}{120}=\frac{3c}{120}=\frac{a}{24}=\frac{b}{15}=\frac{c}{40}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{24}=\frac{b}{15}=\frac{c}{40}=\frac{a}{24}=\frac{2b}{30}=\frac{c}{40}=\frac{a-2b+c}{24-30+40}=\frac{34}{34}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=1.24=24\\b=1.15=15\\c=1.40=40\end{matrix}\right.\)
Vậy...
b)Có: \(3a=7b\Leftrightarrow\frac{a}{7}=\frac{b}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{7}=\frac{b}{3}=\frac{a^2-b^2}{49-9}=\frac{160}{40}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=4.7=28\\b=4.3=12\end{matrix}\right.\)
Vậy...
c) Vì BCNN(15;10;6)=30
\(\Rightarrow15a=10b=6c\Leftrightarrow\frac{15a}{30}=\frac{10b}{30}=\frac{6c}{30}=\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=5k\end{matrix}\right.\)
Thay\(a=2k;b=3k;c=5k\) vào \(abc=-1920\), ta có:
\(2k.3k.5k=-1920\\ \Leftrightarrow30k^3=-1920\\ \Leftrightarrow k^3=-64\\ \Leftrightarrow k^3=\left(-4\right)^3\\ \Leftrightarrow k=-4\)
\(\Rightarrow\left\{{}\begin{matrix}a=-4.2=-8\\b=-4.3=-12\\c=-4.5=-20\end{matrix}\right.\)
Vậy...
Có: 15a=10b=6c
=>\(\frac{a}{10}=\frac{b}{15};\frac{b}{6}=\frac{c}{10}\)
=>\(\frac{a}{20}=\frac{b}{30};\frac{b}{30}=\frac{c}{50}\Rightarrow\)\(\frac{a}{20}=\frac{b}{30}=\frac{c}{50}\)
Đặt \(\frac{a}{20}=\frac{b}{30}=\frac{c}{50}=k\)
=>\(a=20k;b=30k;c=50k\)
Mà abc=-1920
<=>\(20k\cdot30k\cdot50k=-1920\)
\(\Leftrightarrow k^3=-\frac{8}{125}\)
\(\Leftrightarrow k=-\frac{2}{5}\)
\(\Leftrightarrow\begin{cases}a=20k=-8\\b=30k=-12\\c=50k=-20\end{cases}\)