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A=\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
=\(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
=\(\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{72}\)
=\(\left(\frac{14}{15}+\frac{1}{15}\right)-\left(\frac{35}{36}+\frac{1}{36}\right)+\frac{1}{72}\)
=1 - 1 + \(\frac{1}{72}\)= 0 + \(\frac{1}{72}\)= \(\frac{1}{72}\)
A = \(\frac{1}{3}-\frac{3}{4}-\frac{-3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}+\frac{-2}{9}\)
A = \(\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{73}\)
A = \(\left(\frac{3-2}{9}\right)-\left(\frac{27+1}{36}\right)+\left(\frac{9+1}{15}\right)+\frac{1}{73}\)
A = \(\frac{1}{9}-\frac{7}{9}+\frac{6}{9}+\frac{1}{73}\)
A = \(0+\frac{1}{73}=\frac{1}{73}\)
\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)
Ta có: \(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(A=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)-\frac{3}{4}-\frac{1}{36}+\frac{1}{73}\)
\(A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{10}{15}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{2}{3}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{9}+\frac{6}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{7}{9}-\frac{7}{9}+\frac{1}{73}\)
\(A=\frac{1}{73}\)
Vậy: \(A=\frac{1}{73}\)
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