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a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
a) \(\frac{3}{5}+25-\frac{1}{5}=\left(\frac{3}{5}-\frac{1}{5}\right)+25=\frac{2}{5}+25=\frac{2}{5}+\frac{125}{5}=\frac{127}{5}\)
b) \(13\times3\times32,27+67,63\times39=39\times32,27+67,63\times39\)
\(=39\times\left(32,27+67,63\right)\)
\(=39\times99,9=3196,8\)
(Bạn xem lại đề nhé)
c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{99}{100}\)
\(=\frac{1\times2\times3\times...\times99}{2\times3\times4\times....\times100}=\frac{1}{100}\)
a, \(\frac{3}{5}+25-\frac{1}{5}=\frac{127}{5}\)
b, \(\text{13 x 3 x 32,27 + 67,63 x 39 =}3896,1\)
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thank for watching me do homework
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Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)
\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)
\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )
\(\frac{70}{3}-\left(x+4\frac{1}{5}\right)=16\)
\(\Rightarrow x+4\frac{1}{5}=\frac{22}{3}\)
\(\Rightarrow x+\frac{21}{5}=\frac{22}{3}\)
\(\Rightarrow x=\frac{22}{3}-\frac{21}{5}\)
\(\Rightarrow x=\frac{47}{15}\)
b:
3/2 x 4/3 x 5/4 x ......... x 8/7 x 9/8
Ta loai bo so giong nhau o TS va MS
Ta duoc 9/2
Ta có: \(L=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2004}\right)\)
\(\Rightarrow L=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2003}{2004}\)\(=\frac{1.2.3.4....2003}{2.3.4.....2004}=\frac{1}{2004}\)
VẬY \(L=\frac{1}{2004}\)
\(L=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x.....x\left(1-\frac{1}{2003}\right)\)
\(L=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x.....x\frac{2002}{2003}\)
\(L=\frac{1x2x3x4x....x2002}{2x3x4x5x.....x2003}=\frac{1}{2003}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\)
\(=\frac{1}{5}\)