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\(G=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{6}}=\dfrac{1}{5}\)
\(H=\left[\dfrac{32}{99}\cdot\dfrac{14}{9}-\dfrac{25}{99}\right]\cdot\dfrac{11}{83}\)
\(=\dfrac{223}{891}\cdot\dfrac{11}{83}=\dfrac{223}{6723}\)
\(\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\left(1+\dfrac{5}{9}\right)-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{448}{891}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\dfrac{223}{891}.\dfrac{11}{83}\)
\(=\dfrac{223}{6723}\)
\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{2453}{73953}\)
\(\Leftrightarrow H=\dfrac{223}{6723}\)
2) \(A=\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}-\dfrac{1}{6}}{\dfrac{15}{6}+\dfrac{10}{6}-\dfrac{5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3+2-1}{6}}{\dfrac{15+10-5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{4}{6}}{\dfrac{20}{6}}\)
\(\Leftrightarrow A=\dfrac{4}{6}.\dfrac{6}{20}\)
\(\Leftrightarrow A=\dfrac{24}{120}\)
\(\Leftrightarrow A=\dfrac{1}{5}\)
H = [0,(32) . 1,(5) - 0,(25)] . \(\frac{11}{83}\)
H = \(\left(\frac{32}{99}.\frac{15}{9}-\frac{25}{99}\right).\frac{11}{83}\)
H = \(\left(\frac{160}{297}-\frac{75}{297}\right).\frac{11}{83}\)
H = \(\frac{85}{297}.\frac{11}{83}\)
H = \(\frac{85}{2241}\)
H = ( 0 . 5 . 0).11 phần 83 (mk ko bt ghi phân số)
H = 0 . 11
H = 0
Study good