Tính nhanh
0,375-0,3+3/11+3/12 + 1,5+1-0,75
-0,625+0,5-5/11-5/12. 2,5+5/3-1,25
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\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right)\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(=\frac{-3}{5}+\frac{3}{5}=0\)
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
\(A=\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}-\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}-\frac{3}{5}=0\)
Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)
\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)
= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115
= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115
=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{11}\right)}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right)\div\frac{1890}{2005}+115\)
\(A=0\div\frac{1890}{2005}+115\)
\(A=115\)
\(=\left(\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}+\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\right)\cdot\dfrac{2005}{1890}+115\)
\(=\left(\dfrac{3}{5}-\dfrac{3}{5}\right)\cdot\dfrac{2005}{1890}+115\)
=115
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0