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a. \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\)
\(=\frac{1}{3}(\frac{4}{5}+\frac{6}{5})-\frac{5}{3}\)
\(=\frac{1}{3}.2-\frac{5}{3}\)
\(=\frac{2}{3}-\frac{5}{3}\)
\(=-\frac{1}{1}\)
c. \(\frac{6}{7}.\frac{10}{9}+\frac{1}{7}.\frac{10}{9}-\frac{8}{9}\)
\(=\frac{10}{9}\left(\frac{6}{7}+\frac{1}{7}\right)-\frac{8}{9}\)
\(=\frac{10}{9}.1-\frac{9}{8}\)
\(=\frac{10}{9}-\frac{9}{8}\)
\(=-\frac{1}{72}\)
a) -1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - 9 - 10 + 11 + 12 - ... - 2013 - 2014 + 2015 + 2016
= ( -1 - 2 + 3 + 4 ) - ( 5 + 6 - 7 - 8 ) - ( 9 + 10 - 11 - 12 ) - .......... - ( 2013 + 2014 - 2015 - 2016 )
= 4 - ( -4 ) - ( -4 ) - ......... - ( -4 )
= 4 + 4 + 4 +....... + 4
= { [ ( 2016 - 1 ) : 1 + 1 ] : 4 } . 4
= { [ 2015 : 1 + 1 ] : 4 } . 4
= { 2016 : 4 } . 4
= 504 . 4
= 2016
b) \(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):\left(\frac{1}{5}-1\right):.........:\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:\frac{-4}{5}:......:\frac{-99}{100}\)
\(=\frac{-1}{2}.\frac{3}{-2}.\frac{4}{-3}.\frac{5}{-4}.......\frac{100}{-99}\)
\(=\frac{-1.3.4........100}{2.2.3.4......99}\)
\(=\frac{-1.100}{2.2}\)
\(=\frac{-100}{4}\)
\(=-25\)
a) -1-2+3+4-5-6+7+8+...+2016=-3+3-7+7-...-2016+2016=0
b) \(\left(\frac{1}{2}-1\right):...:\left(\frac{1}{100}-1\right)=\frac{-1}{2}:\frac{-2}{3}:\frac{-3}{4}:...:\frac{-99}{100}\)
\(=\)\(\frac{-1}{2}.\frac{-3}{2}.....\frac{-100}{99}=\frac{-1}{2}.\left(-50\right)=25\)
13/50+9/100+41/100+12/50
=(13/50+12/50)+(9/100+41/100)
=1/2+1/2
=1
11) Ta có:
\(\frac{120-0,5.40.5.0,2.20.0,25-20}{1+5+9+...+33+37}\)
\(=\frac{120-\left(0,5.40\right).\left(5.0,2\right).\left(20.0,25\right)-20}{1+5+9+...+33+37}\)
\(=\frac{120-20.1.5-20}{1+5+9+...+33+37}\)
\(=\frac{120-100-20}{1+5+9+...+33+37}\)
\(=\frac{0}{1+5+9+...+33+37}=0\)