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

\(a=\frac{2.9.8+3.12.10+4.15.12+.......+98.297.200}{2.3.4+3.4.5+4.5.6+.........+98.99.100}\)

\(a=\frac{2.\left(3.3\right).\left(4.2\right)+3.\left(4.3\right).\left(5.2\right)+..........+98.\left(99.3\right).\left(100.2\right)}{2.3.4+3.4.5+4.5.6+.................+98.99.100}\)

\(a=\frac{2.3.4.\left(3.2\right)+3.4.5.\left(3.2\right)+............+98.99.100.\left(3.2\right)}{2.3.4+3.4.5+........+98.99.100}\)

\(a=\frac{\left(3.2\right).\left(2.3.4+3.4.5+4.5.6+...........+98.99.100\right)}{2.3.4+3.4.5+4.5.6+............+98.99.100}\)

\(a=3.2\)

\(a=6\)

Vậy a=6.

\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297.200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...98\cdot99\cdot100}\\ =\dfrac{2\cdot3\cdot3\cdot4\cdot2+3\cdot3\cdot4\cdot2\cdot5+4\cdot3\cdot5\cdot2\cdot6+...+98\cdot99\cdot3\cdot100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot2\cdot3+...+98\cdot99\cdot100}\\ =\dfrac{3\cdot2+3\cdot2+6+3\cdot2}{0}=\dfrac{24}{0}=0\)

\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\\ =\dfrac{\left(2\cdot3\right)\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\\ =6\\ a^2=6^2=36\)

\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}=\frac{3.2.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)

a có hoặc ko phải là nhiệm ...

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 \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)

\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2S=\frac{1}{2}-\frac{1}{9900}\)

\(2S=\frac{4949}{9900}\)

\(S=\frac{4949}{19800}\)

Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)

...

\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)

=> 2S = \(\frac{4949}{9900}\)

=> S = \(\frac{4949}{19800}\)

2S=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\)

2S= \(1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)2S= 1- \(\dfrac{1}{100}\)

2S= \(\dfrac{99}{100}\)

S= \(\dfrac{99}{100}.\dfrac{1}{2}\)

S=\(\dfrac{198}{100}\)