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Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}=\dfrac{8}{9}-\left(\dfrac{1}{8.9}+\dfrac{1}{7.8}+\dfrac{1}{6.7}+\dfrac{1}{5.6}+\dfrac{1}{4.5}+\dfrac{1}{3.4}+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)
a)\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}\)
=\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)=1+1=2\)
b) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
=\(\dfrac{8}{9}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
=\(\dfrac{8}{9}-\left(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)
thế này gọi là gian lận nha
tự đăng tự trả lời thì ko nên đăng làm gì
ko đc gì đâu 
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)
\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)
\(=5+\dfrac{19}{4}\)
\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)
\(------\)
\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)
\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)
\(=6-\dfrac{19}{7}\)
\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
câu b, đâu ạ?