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\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)...\left(1+\frac{1}{2015\cdot2017}\right)\)\(A=\frac{1}{2}\left(\frac{1\cdot3+1}{1\cdot3}\right)\left(\frac{2\cdot4+1}{2\cdot4}\right)...\left(\frac{2015\cdot2017+1}{2015\cdot2017}\right)\)
\(A=\frac{1^2}{2}\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\cdot\cdot\frac{2016^2}{2015\cdot2017}\)
\(A=\frac{1^2\cdot2^2\cdot3^2\cdot\cdot\cdot2016^2}{2\cdot1\cdot3\cdot2\cdot4\cdot\cdot\cdot2015\cdot2017}\)
\(A=\frac{2016}{2017}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3+25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3+5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1+7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2}{12}-\frac{5.8}{9}=\frac{1}{6}-\frac{40}{9}=\frac{-77}{18}\)
b ) 3n+2 - 2n+2 + 3n - 2n
= ( 3n+2 + 3n ) - ( 2n+2 + 2n )
= 3n ( 32 + 1 ) - 2n ( 22 + 1 )
= 3n.10 - 2n-1.2.5
= 3n.10 - 2n-1.10
= ( 3n - 2n-1 ).10 chia hết cho 10 ( đpcm )
