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b, ( mik cx ko viết đề bài lun ^_^ )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 13 . 555 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 13 . 5 . 111 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . ( 65 . 111 - 65 . 111 )
= ( 1 + 2 + 3 + ... + 100 ) . ( 12 + 22 + 32 + ... + 102 ) . 0
= 0
Chúc hok tốt !
Mk nhầm sr nhé :
A = \(\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}\)
A = \(\frac{2^{10}.78}{2^{11}.13}\)
A = \(\frac{2^{10}.13.3.2}{2^{11}.13}\)
A = \(\frac{2^{11}.13.3}{2^{11}.13}\)
A = \(3\)
\(\left(\dfrac{1}{2^2}-1\right)\times\left(\dfrac{1}{3^2-1}\right)\times\left(\dfrac{1}{4^2}-1\right)\times...\times\left(\dfrac{1}{100^2}-1\right)\)
\(=\dfrac{3}{2^2}\times\dfrac{8}{3^2}\times\dfrac{15}{4^2}\times...\times\dfrac{100^2-1}{100^2}\)
\(=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{99\times101}{100\times100}\)
\(=\dfrac{1\times2\times3\times...\times99}{2\times3\times4\times...\times100}\times\dfrac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)
\(=\dfrac{1}{100}\times\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
\(\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{3}\cdot...\cdot\dfrac{-9999}{10000}\)
\(=\dfrac{1\cdot\left(-3\right)}{2\cdot2}\cdot\dfrac{2\cdot\left(-4\right)}{3\cdot3}\cdot...\cdot\dfrac{99\cdot\left(-101\right)}{100\cdot100}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{\left(-3\right)\cdot\left(-4\right)\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)
Ở tử số phân số bên phải có số thừa số là: \(101-3+1=99\)
99 là số lẻ nên tử số vế phải sẽ cho ra số âm.
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)
\(=\dfrac{1\cdot\left(-101\right)}{100\cdot2}\)
\(=\dfrac{-101}{200}\)
\(\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{10}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\frac{1}{100}\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot0\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)=0
\(\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).\left(\frac{1}{100}-\left(\frac{1}{2}\right)^2\right)......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)....\left(\frac{1}{100}-\left(\frac{1}{10}\right)^2\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)...\left(\frac{1}{100}-\frac{1}{100}\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).....0......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=0\)