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\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right)\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(=\frac{-3}{5}+\frac{3}{5}=0\)
\(A=\frac{0,375-0,3+\frac{3}{11}+\frac{1}{4}}{0,625+0,5-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}+\frac{1}{10}-\frac{1}{12}-\frac{1}{12}\right)}\)
\(=\frac{3.263.\frac{1}{1320}}{5.67.\frac{1}{1320}}=\frac{789.\frac{1}{1320}}{335,\frac{1}{1320}}=\frac{789}{335}\)
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{11}\right)}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right)\div\frac{1890}{2005}+115\)
\(A=0\div\frac{1890}{2005}+115\)
\(A=115\)
\(A=\frac{0,375-0,3+\frac{3}{10}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(\Rightarrow A=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(\Rightarrow A=\frac{-3.\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(\Rightarrow A=\frac{-3}{5}+\frac{3}{5}\)
\(\Rightarrow A=0\)
Vậy A = 0
@@ Học tốt @@
# Chiyuki Fujito
P = 2005 : ( 3(0,125 - 0,1 + 1/11 + 1/12)/-5(0,125 - 0,1 + 1/11 + 1/12) * 5(0,5 + 1/3 - 0,25)/3(0,5 +1/3 - 0,25) )
P = 2005 : (-3/5 * 5/3)
P = 2005 : (-1) = -2005
Chẳng cần máy tính cũng làm ra, có khi còn nhanh hơn máy tính
\(A=\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}-\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}-\frac{3}{5}=0\)