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\(=\dfrac{1994\left(1994+1\right)-1}{1994^2-1+1994}=\dfrac{1994^2+1993}{1994^2+1993}=1\)
\(\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1993.1995+1995-1}{1993.1995+1994}=1\)
\(\left(7^{1997}-7^{1995}\right):\left(7.7^{1994}\right)\\ =\left(7^{1997}-7^{1995}\right):\left(7^{1+1994}\right)\\ =\left(7^{1997}-7^{1995}\right):7^{1995}\\ =\left(7^{1997}:7^{1995}\right)-\left(7^{1995}:7^{1995}\right)\\ =\left(7^{1997-1995}\right)-1\\ =7^2-1\\ =48\)
a) \(\left(2^{2016}+2^{2017}+2^{2018}\right):\left(2^{2014}+2^{2015}+2^{2016}\right)\)
\(=\dfrac{2^{2016}+2^{2017}+2^{2018}}{2^{2014}+2^{2015}+2^{2016}}\)
\(=\dfrac{2^{2016}\left(1+2+2^2\right)}{2^{2014}\left(1+2+2^2\right)}\)
\(=\dfrac{2^{2016}}{2^{2014}}\)
\(=2^{2016-2014}\)
\(=2^2\)
\(=4\)
b)
\(3^{500}=3^{5.100}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=7^{3.100}=\left(7^3\right)^{100}=343^{100}\)
Vì \(243< 343\)
Nên \(243^{100}< 343^{100}\)
Vậy \(3^{500}< 7^{300}\)
tthấy cách này dễ hơn :
(22016+22017+22018):(22014+22015+22016)
=22016.(1+2+22):22014.(1+2+22)
=(22016.7)+(22014.7)
=22
=4
\(A=x^2-y^2-2y-1\)
\(=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
\(=\left(93-6-1\right)\left(93+6+1\right)=86\cdot100=8600\)
B k hiểu đề là j
\(S=\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}-\dfrac{3}{5\cdot9}-\dfrac{3}{9\cdot13}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}\right)-\dfrac{3}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\right)-\dfrac{3}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}\right)\)
\(=\dfrac{2}{5}\cdot\dfrac{3}{20}-\dfrac{3}{4}\cdot\dfrac{8}{65}=\dfrac{-21}{650}\)