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binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
Xem thêm tại: http://loigiaihay.com/bai-77-trang-39-phan-so-hoc-sgk-toan-6-tap-2-c41a5943.html#ixzz4eU1fQCGw
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-16}{10}:\dfrac{5}{3}\)
\(A=\dfrac{-8}{5}.\dfrac{3}{5}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{14}{10}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\left(1+\dfrac{2}{3}\right)\)
\(A=\dfrac{-8}{5}:\dfrac{5}{3}\)
\(A=\dfrac{-24}{25}\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
\(B=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}:\dfrac{11}{5}\)
\(B=\dfrac{3}{7}-\dfrac{2}{3}\)
\(B=\dfrac{-5}{21}\)
B = \(\dfrac{3}{5}+\dfrac{3}{5^2}+\dfrac{3}{5^3}+...+\dfrac{3}{5^{2016}}\)
=> 5B = \(3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^{2015}}\)
=> 4B = \(3-\dfrac{3}{5^{2016}}\)
=> B = \(\dfrac{3-\dfrac{3}{5^{2016}}}{4}\)
@Nguyễn Đình Dũng có thể đưa ra kết quả chính xác được không?