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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 = \(\frac{100}{51}\)
$D=\dfrac{\left(1-\dfrac12\right)\left(1-\dfrac13\right)\cdots\left(1-\dfrac1{25}\right)}{\left(1+\dfrac12\right)\left(1+\dfrac13\right)\cdots\left(1+\dfrac1{25}\right)}$
$=\dfrac{\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{24}{25}}{\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{26}{25}}$
$=\dfrac{\dfrac1{25}}{\dfrac{26}{2}}$
$=\dfrac1{25}\cdot\dfrac2{26}$
$=\dfrac1{325}.$
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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
A=\(\frac{1,25x10x4x24,4x2}{6,1x2x6,25x4x2}\)
Thanh Huyền trả lời ít thế ai hiểu ?