\(2,\\ a,A=-\dfrac{8}{15}\cdot\dfrac{5}{4}+\left(\dfrac{53}{10}-\dfrac{23}{6}\right):\dfrac{4}{9}\\ A=-\dfrac{2}{3}+\dfrac{22}{15}\cdot\dfrac{9}{4}=\dfrac{33}{10}-\dfrac{2}{3}=\dfrac{79}{30}\\ b,B=4\cdot\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{4}:\dfrac{1}{4}+1\\ B=\dfrac{1}{2}+\dfrac{1}{6}+1+1=\dfrac{2}{3}+1+1=\dfrac{8}{3}\)
\(3,\)
\(a,\)Áp dụng t/c dtsbn:
\(\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{2b}{10}=\dfrac{a-b+c}{4}=\dfrac{a-2b+c}{-1}\\ \Rightarrow\dfrac{a-b+c}{a-2b+c}=\dfrac{4}{-1}=-4\)
\(b,\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{3x-3}{15}=\dfrac{5y-10}{15}=\dfrac{6z-12}{12}\)
Áp dụng t/c dtsbn:
\(\dfrac{x-1}{5}=\dfrac{y-2}{3}=\dfrac{z-2}{2}=\dfrac{3x-3}{15}=\dfrac{5y-10}{15}=\dfrac{6z-12}{12}\\ =\dfrac{3x-3-5y+10+6z-12}{15-15+12}=\dfrac{\left(3x-5y+6x\right)-3+10-12}{12}=\dfrac{4}{12}=3\\ \Rightarrow\left\{{}\begin{matrix}x-1=15\\y-2=9\\z-2=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=16\\y=11\\z=8\end{matrix}\right.\)
