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1998.1998 – 2000.1996 = 1998.(1996 + 2) – (1998 + 2).1996
= (1998.1996 + 2.1998) – (1998.1996 + 2.1996)
= 2.1998 – 2.1996 = 2(1998 – 1996) = 2.2 = 4
1. 1998.1998 – 2000.1996 = 1998.(1996 + 2) – (1998 + 2).1996
= (1998.1996 + 2.1998) – (1998.1996 + 2.1996)
= 2.1998 – 2.1996 = 2(1998 – 1996) = 2.2 = 4
\(\frac{2^8\times6}{3^3\times5^4}\div\frac{8^3\times9}{5^3\times3^3}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-4^2\right)\)
\(=\frac{2^9\times3}{3^3\times5^4}\times\frac{5^3\times3^3}{2^9\times3^2}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-2^4\right)\)
\(=\frac{2^9\times3^4\times5^3}{3^5\times5^4\times2^9}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\times0\)
\(=\frac{1}{3\times5}-0\)
\(=\frac{1}{15}\)
Ta có: \(A=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+......+\frac{5}{64.69}\right)\)
\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+......+\frac{1}{64}-\frac{1}{69}\right)\)
\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{69}\right)=\frac{1}{5}\times\frac{65}{276}\)
\(\Rightarrow A=\frac{13}{276}\)
Vậy \(A=\frac{13}{276}\)
1) 1998.1998 - 2000.1996
= 1998.1998 - 1998.1996 - 2.1996
= 2.1998 - 2.1996
= 2.2
= 4
2)
Bài 1: 1998 x 1998-2000 x 1996
=1998x(2000-2) - 2000 x 1996
=1998x2000 - 1998x2 - 2000x1996
=( 1998x2000- 2000x1996) - 1998 x 2
= 2000x( 1998-1996 ) - 1998 x 2
= 2000 x 2 - 1998 x 2
=2 x (2000-1998)
= 2 x 2
= 4