Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
A = \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{2004}{2004}-\frac{1}{2004}\right)\)
A = \(\frac{1}{2}\)x\(\frac{2}{3}.\)\(\frac{3}{4}....\)\(\frac{2003}{2004}\)
A = \(\frac{1}{2004}\)

G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)

đụ cha mi
mi trù ta thi rớt HK II mà ta giúp mày hả
mấy bài này cũng dễ ẹt nữa
đừng có mơ ta sẽ giúp mày
ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)

\(3^2\times\frac{1}{243}\times81^2\times\frac{1}{3^3}\)
\(=3^2\times\frac{1}{3^5}\times\left(3^4\right)^2\times\frac{1}{3^3}\)
\(=\left(3^2\times3^8\right)\times\left(\frac{1}{3^5}\times\frac{1}{3^3}\right)\)
\(=3^{10}\times\frac{1}{3^8}\)
\(=3^2\)
\(=9\)
\(\left(4\times2^5\right)\div\left(2^3\times\frac{1}{6}\right)\)
\(=\left(2^2\times2^5\right)\div\left(2^3\times\frac{1}{2\times3}\right)\)
\(=2^7\div2^2\times3\)
\(=2^5\times3\)
\(=96\)
\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
\(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}\)
\(=\left(3^2.3^8\right).\left(\frac{1}{3^5}.\frac{1}{3^3}\right)\)
\(=3^{10}.3^{-8}\)
\(=3^2=9\)
\(\left(4.2^5\right):\left(2^3.\frac{1}{6}\right)\)
\(=2^7:2^2.3\)
\(=2^5.3\)
\(=96\)
\(A=4+\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{19}\right)\cdot\left(1-\frac{1}{20}\right)\)
\(A=4+\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\right)\)
\(A=4+\frac{1\cdot2\cdot3\cdot...\cdot18\cdot19}{2\cdot3\cdot4\cdot...\cdot19\cdot20}\)
\(A=4+\frac{1}{20}\)
\(A=\frac{81}{20}\)
Bạn có thể giải rõ vì sao bằng 4 ko ? Giải được mình k