S=3+3/2+3/2²+......+3/2^9
A=2.9.8+3.12.10+4.15.12+....+98.297.200/23.4+3.4.5+4.5.6+…98.99.100 tinh a²
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LN
5
24 tháng 4 2017
Ta có:
\(2.9.8+3.12.10+...+98.297.200\)
\(=2.3.4.3.2+3.4.5.3.2+...+98.99.100.3.2\)
\(=6.\left(2.3.4+3.4.5+...+98.99.100\right)\)
Thế lại bài toán (sửa đề luôn)
\(a=\frac{2.9.8+3.12.10+...+98.297.200}{2.3.4+3.4.5+...+98.99.100}\)
\(=\frac{6.\left(2.3.4+3.4.5+...+98.99.100\right)}{2.3.4+3.4.5+...+98.99.100}=6\)
\(\Rightarrow a^2=6^2=36\)
NN
1
5 tháng 3 2017
\(a=\frac{2.9.8+3.12.10+4.15.12+.......+98.297.200}{2.3.4+3.4.5+4.5.6+.........+98.99.100}\)
\(a=\frac{2.\left(3.3\right).\left(4.2\right)+3.\left(4.3\right).\left(5.2\right)+..........+98.\left(99.3\right).\left(100.2\right)}{2.3.4+3.4.5+4.5.6+.................+98.99.100}\)
\(a=\frac{2.3.4.\left(3.2\right)+3.4.5.\left(3.2\right)+............+98.99.100.\left(3.2\right)}{2.3.4+3.4.5+........+98.99.100}\)
\(a=\frac{\left(3.2\right).\left(2.3.4+3.4.5+4.5.6+...........+98.99.100\right)}{2.3.4+3.4.5+4.5.6+............+98.99.100}\)
\(a=3.2\)
\(a=6\)
Vậy a=6.
NH
2
20 tháng 4 2018
\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297.200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...98\cdot99\cdot100}\\ =\dfrac{2\cdot3\cdot3\cdot4\cdot2+3\cdot3\cdot4\cdot2\cdot5+4\cdot3\cdot5\cdot2\cdot6+...+98\cdot99\cdot3\cdot100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot2\cdot3+...+98\cdot99\cdot100}\\ =\dfrac{3\cdot2+3\cdot2+6+3\cdot2}{0}=\dfrac{24}{0}=0\)
20 tháng 4 2018
\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\\ =\dfrac{\left(2\cdot3\right)\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\\ =6\\ a^2=6^2=36\)
18 tháng 4 2016
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}=\frac{3.2.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
21 tháng 5 2018
Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
HT
18 tháng 8 2015
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{98.99.100}\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
SAI ĐỀ RỒI BẠN. SỬA 23=2.3
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)
\(\frac{1.2.3.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{\left(2.3.4+3.4.5+4.5.6+...+98.99.10\right)}\)
\(=6\)
VẬYa2=62=36