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=> A= \(\frac{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}\right).23.7.1009}{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right).23.7.1009}\) + \(\frac{1}{30.1009-160}\)
=> A= \(\frac{7.1009+23.1009-23.7}{7.1009+23.1009-23.7+1}\) + \(\frac{1}{7.1009+23.1009-23.7+1}\) = \(\frac{7.1009+23.1009-23.7+1}{7.1009+23.1009-23.7+1}\) = 1.
\(B=\frac{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}\right).23.7.1009}{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{23}.\frac{1}{7}.\frac{1}{1009}\right)}+\frac{1}{\left(23+7\right).1009-161+1}\)
\(=\frac{7.1009+23.1009-23.7}{7.1009+23.1009-23.7+1}+\frac{1}{7.1009+23.1009-23.7+1}=1\)
\(\Rightarrow B=1\)
\(A=\left(\frac{1}{9}+\frac{1}{23}-\frac{1}{2009}\right):\left(\frac{1}{23}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right)+1:\left(30.1009-160\right)\)
sAI ĐỀ RỒI BẠN
a,
A = 20102010.[710:78-3.16-22010:22010]
= 20102010.[72-48-1]
= 20102010.0 = 0
b,
B = 1
\(A=2010^{2010}.\left[7^{10}:7^8-3.16-2^{2010}:2^{2010}\right]\)
\(A=2010^{2010}.\left[7^2-48-1\right]\)
\(A=2010^{2010}.0\)
\(Vay\)\(A=0\)