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\(A=13.15+15.17+17.19+...+99.101\)
\(\Rightarrow6A=13.15.6+15.17.6+17.19.6+...+99.101.6\)
\(\Rightarrow6A=13.15.\left(17-11\right)+15.17.\left(19-13\right)+17.19.\left(21-15\right)+...+99.101.\left(103-97\right)\)
\(\Rightarrow6A=\left(13.15.17-11.13.15\right)+\left(15.17.19-13.15.17\right)+\left(17.19.21-15.17.19\right)+...+\left(99.101.103-97.99.101\right)\)
\(\Rightarrow6A=99.101.103-11.13.15\)
\(\Rightarrow6A=1027752\)
\(\Rightarrow A=171292\)
=4x(\(\frac{1}{11x13}\)+\(\frac{1}{13x15}\)+.......+\(\frac{1}{99x101}\))
=4x(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\))
4x(\(\frac{1}{11}\)-\(\frac{1}{101}\))
=4x \(\frac{90}{1111}\)
=\(\frac{360}{1111}\)
\(\frac{4}{11\times13}+\frac{4}{13\times15}+\frac{4}{15\times17}+...+\frac{4}{99\times101}\)
\(=\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+\frac{4}{15}-\frac{4}{17}+...+\frac{4}{99}-\frac{4}{101}\)
\(=\frac{4}{11}-\frac{4}{101}\)
\(=\frac{360}{1111}\)
Ta có : B = 1.1 + 2.2 + 3.3 + ... + 1999.1999
= 1.(2 - 1) + 2.(3 - 1) + 3(4 - 1) + ... + 1999(2000 - 1)
= (1.2 + 2.3 + 3.4 + ... + 1999.2000) - (1 + 2 + 3 + .... + 1999)
Đặt A = 1.2 + 2.3 + 3.4 + ... + 1999.2000
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 1999.2000.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + 1999.2000.(2001 - 1998)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 1999.2000.2001 - 1998.1999.2000
= 1999.2000.2001
=> A = 1999.2000.2001/3
Khi đó B = A - (1 + 2 + 3 + .... + 1999)
= 1999.2000.2001/3 - 1999.(1999 + 1)/2
= 1999.2000.667 - 1999.1000
= 1999.(2000.667 - 1000)
= 1999 . 1 333 000
Vậy B = 1999 . 1333000
\(D=\left(4\cdot25\right)\cdot\left(8\cdot125\right)\cdot\left(5\cdot42\right)\cdot2\)
\(D=100\cdot1000\cdot210\cdot2\)
\(D=100\cdot1000\cdot420\)
\(D=100000\cdot420\)
\(D=42000000\)
D=2.4.5.8.25.42.125
D=(2.5)(4.25)(8.125).42
D=10.100.1000.42
D=42000000