Dễ thế cũng hỏi ở trường tính hay lắm mà:
a) \(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}\)
\(=\dfrac{3}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=\dfrac{3}{7}.\left(\dfrac{8+5-2}{11}\right)=\dfrac{3}{7}.\dfrac{11}{11}=\dfrac{3}{7}\)
b) \(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}\)
\(=\dfrac{3}{13}.\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)\)
\(=\dfrac{3}{13}.\left(\dfrac{8+11+6}{25}-1=\right)\dfrac{3}{13}.0=0\)
a,
\(\dfrac{3}{7}.\dfrac{8}{11}+\dfrac{3}{7}.\dfrac{5}{11}-\dfrac{3}{7}.\dfrac{2}{11}=\dfrac{3}{7}\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)=\dfrac{3}{7}.1=\dfrac{3}{7}\)
b,
\(\dfrac{3}{13}.\dfrac{8}{25}+\dfrac{3}{13}.\dfrac{11}{25}+\dfrac{3}{13}.\dfrac{6}{25}-\dfrac{3}{13}=\dfrac{3}{13}\left(\dfrac{8}{25}+\dfrac{11}{25}+\dfrac{6}{25}-1\right)=\dfrac{3}{13}.0=0\)