\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)

=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)

=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)

Đáp số: C=1

C=1

HT

C = 1

JY 

\(-1\frac{1}{2}.\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{1999}\right)\)

\(-\frac{3}{2}.\left(\frac{-4}{3}\right).\left(-\frac{5}{4}\right)....\left(-\frac{2000}{1999}\right)=-\frac{3.4.5...2000}{2.3.4...1999}=-1000\)

\(A=\frac{\frac{2000\cdot2001\cdot2002\cdot...\cdot2999}{1\cdot2\cdot3\cdot...\cdot1000}}{\frac{1001\cdot1002\cdot1003\cdot...\cdot2999}{1\cdot2\cdot3\cdot...\cdot1999}}=\frac{2000\cdot2001\cdot2002\cdot...\cdot2999}{1\cdot2\cdot3\cdot...\cdot1000}\times\frac{1\cdot2\cdot3\cdot...\cdot1999}{1001\cdot1002\cdot1003\cdot...\cdot2999}\)

\(A=1\)

\(=\frac{-1}{2}.\frac{-2}{3}......................\frac{-1998}{1999}.\frac{-1999}{2000}\)

\(=\frac{\left(-1\right).\left(-2\right)....................\left(-1999\right)}{1.2.3........................2000}\)

\(=\frac{-1}{2000}\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1998}{1999}.\frac{1999}{2000}=\frac{1}{2000}\)

duyệt đi

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1999}\right).\left(1-\frac{1}{2000}\right)\)

=\(\vec{\frac{1.2.3...1998.1999}{2.3.4...1999.2000}}\)\(=\frac{1}{2000}\)

\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)

\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)

\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)

\(=\frac{1}{2.2-2.2+2.2}\)

\(=\frac{1}{2.2}=\frac{1}{4}\)

Giúp mik với

trước 5h nha

a) \(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+1995}\)

\(=\frac{120-\left[\left(-0,5\right).\left(-0,2\right)\right].\left[\left(-40\right).0,25\right].\left[\left(-5\right).\left(20\right)\right]}{\left(1995+5\right).\left[\left(1995-5\right)\div5+1\right]\div2}\)

\(=\frac{120-0,1.\left(-10\right).\left(-100\right)}{2000.399\div2}\)

\(=\frac{120-100}{1000.399}\)

\(=\frac{1}{19950}\)

b) \(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)

\(=\frac{5.18-2.5.27+3.5.36}{10.2.18-20.2.27+5.2.3.2.36}\)

\(=\frac{5.18-2.5.27+3.5.36}{20.18-20.2.27+20.3.36}\)

\(=\frac{5.\left(18-2.27+3.36\right)}{20.\left(18-2.27+3.36\right)}\)

\(=\frac{1}{4}\)

c) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)

\(=\left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right).\left(\frac{-3}{4}\right)...\left(\frac{-1998}{1999}\right)\)

\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-1998\right)}{2.3.4...1999}\)

\(=\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)

Ta có : 1998 số (-1) mà 1998 là số chẵn

Vậy tích của 1998 số (-1) = 1

\(\Rightarrow\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)

\(=\frac{1}{1999}\)

a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)

a) =3/2 . 4/3 . 5/4 ...100/99

   =\(\frac{3.4.5...100}{2.3.4..99}\)

  =\(\frac{100}{2}\)

b) =

b) = \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=-1\left(\frac{1.2.3...99}{2.3.4...100}\right)=-\frac{1}{100}\)