Question

[Hoang Khanh Linh]

Áp dụng tính chất các phép tính và quy tắc dấu ngoặc để tính giá trị biểu thức sau :

Answer 1

[Quìn]

\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(A=11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)

\(A=\left(11-2-5\right)+\left(\dfrac{3}{13}-\dfrac{4}{7}-\dfrac{3}{13}\right)\)

\(A=4+\dfrac{-4}{7}\)

\(A=\dfrac{24}{7}=3\dfrac{3}{7}\)

\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(B=\left[\left(6+3\right)+\left(\dfrac{4}{9}+\dfrac{7}{11}\right)\right]-4\dfrac{4}{9}\)

\(B=\left[9+\dfrac{107}{99}\right]-4\dfrac{4}{9}\)

\(B=10\dfrac{71}{99}-4\dfrac{4}{9}\)

\(B=10-4+\dfrac{71}{99}-\dfrac{4}{9}\)

\(B=6\dfrac{3}{11}\)

Answer 2

[Quìn]

\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.1+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}+1+\dfrac{5}{7}\)

\(C=0+1\)

\(C=1\)

\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(D=\left(\dfrac{3}{8}.\dfrac{8}{3}\right).\left(\dfrac{7}{10}.\dfrac{5}{28}\right).20\)

\(D=1.\dfrac{1}{8}.20\)

\(D=\dfrac{5}{2}\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)

\(\Rightarrow E=0\)