Bài 1:
\(1,=-\dfrac{4}{3}\\ 2,=\dfrac{1}{3}\left(5\dfrac{2}{5}+1\dfrac{2}{5}\right)=\dfrac{1}{3}\cdot6\dfrac{4}{5}=\dfrac{1}{3}\cdot\dfrac{34}{5}=\dfrac{34}{15}\\ 3,=\dfrac{2}{3}\cdot\dfrac{4}{7}+\dfrac{2}{3}\cdot\dfrac{3}{7}=\dfrac{2}{3}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)=\dfrac{2}{3}\cdot1=\dfrac{2}{3}\\ 4,=\dfrac{5}{11}\left(-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{1}{3}\right)=\dfrac{5}{11}\left(1-1\right)=0\)
Bài 2:
\(1,\Rightarrow\dfrac{5}{3}x=-\dfrac{13}{21}\Rightarrow x=-\dfrac{13}{35}\\ 2,\Rightarrow\dfrac{2}{3}:x=-\dfrac{22}{3}\Rightarrow x=-\dfrac{1}{11}\\ 3,\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{15}\)
Bài 3:
Áp dụng tc dtsbn:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{45}{9}=5\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
Bài 4:
Gọi số cây 6A,6B,6C trồng đc lần lượt là a,b,c(cây;a,b,c∈N*)
Áp dụng tc dstbn:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a+b+c}{2+3+5}=\dfrac{80}{10}=8\\ \Rightarrow\left\{{}\begin{matrix}a=16\\b=24\\c=40\end{matrix}\right.\)
Vậy ...
Bài 3.
\(\dfrac{x}{4}=\dfrac{y}{5};x+y=45\)
Theo tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{45}{9}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\cdot4=20\\y=5\cdot5=25\end{matrix}\right.\)
