1/ Gọi số học sinh 3 lớp 7A; 7B; 7C lần lượt là \(a,b,c\)
Theo bài ta có :
\(a+b+c=93\left(hs\right)\)
\(2a=3b=5c\)
\(\Leftrightarrow\dfrac{2a}{30}=\dfrac{3b}{30}=\dfrac{5c}{30}\)
\(\Leftrightarrow\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{6}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{6}=\dfrac{a+b+c}{15+10+6}=\dfrac{93}{31}=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{15}=3\\\dfrac{b}{10}=3\\\dfrac{c}{6}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=45\\b=30\\c=18\end{matrix}\right.\)
Vậy ...
b/ \(\dfrac{1}{5}+\dfrac{7}{13}-\dfrac{3}{15}+\dfrac{6}{13}-1\)
\(=\dfrac{1}{5}+\dfrac{7}{13}-\dfrac{1}{5}+\dfrac{6}{13}-1\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{7}{13}+\dfrac{6}{13}\right)-1\)
\(=0+1-1=0\)
Câu 2
\(\dfrac{1}{5}+\dfrac{7}{13}-\dfrac{3}{15}+\dfrac{6}{13}-1=\left(\dfrac{1}{5}-\dfrac{3}{15}\right)+\left(\dfrac{7}{13}+\dfrac{6}{13}\right)-1\)
=0+1-1=0
