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0,5+0,(3)+0,1(6)/2,5+0,1(6)+0,8(3)
=0,5+1/3+1/6/2,5+1/6+5/6
=1/3,5=2/7
k mk nha
Đặt GTBT là A, ta có:
\(A=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}=\frac{1}{5}\)
\(G=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)\(=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}\)\(=\frac{1}{5}.\left(\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}\right)\)\(=\frac{1}{5}.1=\frac{1}{5}\)
\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{2453}{73953}\)
\(\Leftrightarrow H=\dfrac{223}{6723}\)
2) \(A=\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}-\dfrac{1}{6}}{\dfrac{15}{6}+\dfrac{10}{6}-\dfrac{5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3+2-1}{6}}{\dfrac{15+10-5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{4}{6}}{\dfrac{20}{6}}\)
\(\Leftrightarrow A=\dfrac{4}{6}.\dfrac{6}{20}\)
\(\Leftrightarrow A=\dfrac{24}{120}\)
\(\Leftrightarrow A=\dfrac{1}{5}\)
\(\frac{0,5+\frac{1}{3}-\frac{1}{6}}{2,5+\frac{5}{3}-\frac{5}{6}}\)
= \(\frac{\frac{3}{6}+\frac{2}{6}-\frac{1}{6}}{\frac{15}{6}+\frac{10}{6}-\frac{5}{6}}\)
= \(\frac{\frac{4}{6}}{\frac{20}{6}}\)
= \(\frac{4}{6}:\frac{20}{6}\)
= \(\frac{2}{3}.\frac{3}{10}\)
= \(\frac{1}{5}\)