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Câu hỏi: Tính: a) $\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}$
b) $\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}$
c) $\dfrac{3}{5}:\left(\dfrac{1}{4} \cdot \dfrac{7}{5}\right)$
d) $\dfrac{10}{11}+\dfrac{4}{11}: 4-\dfrac{1}{8}$
b) $\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}$
c) $\dfrac{3}{5}:\left(\dfrac{1}{4} \cdot \dfrac{7}{5}\right)$
d) $\dfrac{10}{11}+\dfrac{4}{11}: 4-\dfrac{1}{8}$
Phương pháp giải
Thực hiện đúng thứ tự thực hiện phép tính trong ngoặc trước --> nhân, chia --> cộng, trừ
Lời giải chi tiết
a) $\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{4}{6}=\dfrac{2}{3}$
b) $\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{31}{24}$
c) $\dfrac{3}{5}:\left(\dfrac{1}{4} \cdot \dfrac{7}{5}\right)=\dfrac{3}{5}: \dfrac{7}{20}=\dfrac{3}{5} \cdot \dfrac{20}{7}=\dfrac{12}{7}$
d) $\dfrac{10}{11}+\dfrac{4}{11}: 4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11} \cdot \dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}$ $=1-\dfrac{1}{8}=\dfrac{7}{8}$
Thực hiện đúng thứ tự thực hiện phép tính trong ngoặc trước --> nhân, chia --> cộng, trừ
Lời giải chi tiết
a) $\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{4}{6}=\dfrac{2}{3}$
b) $\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{31}{24}$
c) $\dfrac{3}{5}:\left(\dfrac{1}{4} \cdot \dfrac{7}{5}\right)=\dfrac{3}{5}: \dfrac{7}{20}=\dfrac{3}{5} \cdot \dfrac{20}{7}=\dfrac{12}{7}$
d) $\dfrac{10}{11}+\dfrac{4}{11}: 4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11} \cdot \dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}$ $=1-\dfrac{1}{8}=\dfrac{7}{8}$