Câu 3:
Áp dụng tính chất của dãy tỉ số bằg nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{3}=\dfrac{z}{8}=\dfrac{x-y+z}{9-3+8}=\dfrac{56}{14}=4\)
Do đó: x=36; y=12; z=32
\(Bài1\\ a,=\left(\dfrac{11}{19}+\dfrac{8}{19}\right)+\left(\dfrac{19}{18}-\dfrac{1}{18}\right)+5,2=\dfrac{19}{19}+\dfrac{18}{18}+5,2=1+1+5,2=7,2\\ b,=\dfrac{3}{7}\left(\dfrac{16}{15}-\dfrac{2}{15}\right)=\dfrac{3}{7}.\dfrac{14}{15}=\dfrac{2}{5}\\ c,=\left(\dfrac{5}{12}:\dfrac{10}{3}\right)+\left(\dfrac{4-3}{6}\right)^2=\dfrac{5}{12}.\dfrac{3}{10}+\left(\dfrac{1}{6}\right)^2=\dfrac{1}{8}+\dfrac{1}{36}=\dfrac{36+8}{288}=\dfrac{11}{72}\)
