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a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
5, \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)
= \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}\).1
=\(\dfrac{1}{7}\)
5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\) = \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}.\dfrac{7}{5}=\dfrac{1}{5}\)
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
B= (7 \(\dfrac{4}{9}\) + 3\(\dfrac{8}{13}\)) - 5\(\dfrac{4}{9}\)
B=(\(\dfrac{67}{9}\)+ \(\dfrac{47}{13}\)) - \(\dfrac{49}{9}\)
B=\(\dfrac{116}{9}\)-\(\dfrac{47}{13}\)
B=\(\dfrac{116-47}{117}\)
B=\(\dfrac{23}{39}\)
tui cũng không biết đúng hông nữa
Đề bài : Áp dụng tính chất các phép tính và quy tắc dấu ngoặc để tính giá trị các biểu thức
G=\(\dfrac{\left(-2\right)}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{2}{3}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)}{3}+\dfrac{2}{3}+\dfrac{\left(-5\right)}{7}+\dfrac{\left(-2\right)}{7}\)
\(\Rightarrow G=\dfrac{\left(-2\right)+2}{3}+\dfrac{\left(-5\right)+\left(-2\right)}{7}\)
\(\Rightarrow G=0+\dfrac{-7}{7}\)
\(\Rightarrow G=-1\)
\(H=\dfrac{\left(-5\right)}{7}\cdot\dfrac{2}{11}+\dfrac{\left(-5\right)}{7}\cdot\dfrac{9}{11}\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot\left(\dfrac{2+9}{11}\right)\)
\(\Rightarrow H=\dfrac{\left(-5\right)}{7}\cdot1\)
\(\Rightarrow H=\dfrac{-5}{7}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
\(\dfrac{-5}{ 7} \) .\(\dfrac{5}{11}+\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{4}{11}\)
=\(\dfrac{-5}{7}.(\dfrac{5}{11}+\dfrac{2}{11}+\dfrac{4}{11})\)
=\(\dfrac{-5}{7}.1 \)
=\(\dfrac{-5}{7}\)