a+b+c=? biet a=5, b=4 ,c=6
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Áp dụng tính chất của dãy các tỉ số bằng nhau, ta có:
\(\frac{a}{4}=\frac{b}{6}=\frac{c}{5}=\frac{a+b+c}{4+5+6}=\frac{100}{15}=\frac{20}{3}\)
\(\Rightarrow a=\frac{20}{3}.4=\frac{80}{3}\)
\(b=\frac{20}{3}.6=40\)
\(c=\frac{20}{3}.5=\frac{100}{3}\)
\(a:4=b:5=c:6\)
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{100}{15}=\frac{20}{3}\)
\(\Leftrightarrow\frac{a}{4}=\frac{20}{3}\)\(\Rightarrow a=\frac{4.20}{3}=\frac{80}{3}\)
\(\Leftrightarrow\frac{b}{5}=\frac{20}{3}\Rightarrow b=\frac{5.20}{3}=\frac{100}{3}\)
\(\Leftrightarrow\frac{c}{6}=\frac{20}{3}\Rightarrow c=\frac{6.20}{3}=40\)
Vậy \(a=\frac{80}{3};b=\frac{100}{3};c=40\)
Taa co : a:b:c:d=7:6:5:4 va a+b+c+d=66
\(a:b:c:d=7:6:5:4=\frac{a}{7}=\frac{b}{6}=\frac{c}{5}=\frac{d}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{7}=\frac{b}{6}=\frac{c}{5}=\frac{d}{4}=\frac{a+b+c+d}{7+6+5+4}=\frac{66}{22}=3\)
Suy ra : \(\frac{a}{7}=3\Rightarrow a=3.7=21\)
\(\frac{b}{6}=3\Rightarrow b=3.6=18\)
\(\frac{c}{5}=3\Rightarrow c=3.5=15\)
\(\frac{d}{4}=3\Rightarrow d=3.4=12\)
a,
\(a+b=-9\\ b+c=2\\ c+a=-3\\ \Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\\ 2a+2b+2c=-10\\ 2\left(a+b+c\right)=-10\\ a+b+c=-5\\ a+b=-9\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-9\right)+c=-5\Rightarrow c=4\\ b+c=2\\ \Rightarrow a+b+c=-5\Leftrightarrow a+2=-5\Rightarrow a=-7\\ c+a=-3\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-3\right)+b=-5\Rightarrow b=-2\)
Vậy \(a=-7;b=-2;c=5\)
b,
\(a+b=\dfrac{1}{2}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{-5}{6}\\ \Rightarrow a+b+b+c+c+a=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{-5}{6}\\ 2a+2b+2c=\dfrac{6}{12}+\dfrac{9}{12}+\dfrac{-10}{12}\\ 2\left(a+b+c\right)=\dfrac{5}{12}\\ a+b+c=\dfrac{5}{24}\\ a+b=\dfrac{1}{2}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{2}+c=\dfrac{5}{24}\Rightarrow c=\dfrac{-7}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{5}{24}\Rightarrow a=\dfrac{-13}{24}\\ a+c=\dfrac{-5}{6}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow b+\dfrac{-5}{6}=\dfrac{5}{24}\Rightarrow b=\dfrac{25}{24}\)
Vậy \(a=\dfrac{-13}{24};b=\dfrac{25}{24};c=\dfrac{-7}{24}\)
c,
\(a+b=2\\ b+c=6\\ c+a=3\\ \Rightarrow a+b+b+c+c+a=2+6+3\\ 2a+2b+2c=11\\ 2\left(a+b+c\right)=11\\ a+b+c=5,5\\ a+b=2\\ \Rightarrow a+b+c=5,5\Leftrightarrow2+c=5,5\Rightarrow c=3,5\\ b+c=6\\ \Rightarrow a+b+c=5,5\Leftrightarrow a+6=5,5\Rightarrow a=-0,5\\ c+a=3\\ \Rightarrow a+b+c=5,5\Leftrightarrow b+3=5,5\Rightarrow b=2,5\)
Vậy \(a=-0,5;b=2,5;c=3,5\)
d,
\(a+b=\dfrac{5}{6}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+b+c+c+a=\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{5}{3}\\ 2a+2b+2c=\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{20}{12}\\ 2\left(a+b+c\right)=\dfrac{13}{4}\\ a+b+c=\dfrac{13}{8}\\ a+b=\dfrac{5}{6}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow\dfrac{5}{6}+c=\dfrac{13}{8}\Rightarrow c=\dfrac{19}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{13}{8}\Rightarrow a=\dfrac{7}{8}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow b+\dfrac{5}{3}=\dfrac{13}{8}\Rightarrow b=\dfrac{-1}{24}\)
Vậy \(a=\dfrac{7}{8};b=\dfrac{-1}{24};c=\dfrac{19}{24}\)
\(\left\{{}\begin{matrix}a+b=-9\\b+c=2\\c+a=-3\end{matrix}\right.\)
\(\Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\)
\(\Rightarrow2a+2b+2c=-10\)
\(\Rightarrow2\left(a+b+c\right)=-10\)
\(\Rightarrow a+b+c=-5\)
\(\Rightarrow\left\{{}\begin{matrix}c=-5-9=-14\\a=-5-2=-7\\b=-5-\left(-3\right)=-2\end{matrix}\right.\)
a + b + c
= 5 + 4 + 6
= 5 + 10
= 15
a+b+c=5+4+6=15