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#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
\(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{13}{28}\)
\(B=\frac{13}{28}.7=\frac{13}{4}\)
\(B=\frac{5}{2}+\left(\frac{4}{1.11}+\frac{3}{11.2}\right)+\left(\frac{1}{2.15}+\frac{13}{15.4}\right)\)
\(B=\frac{5}{2}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}\left(\frac{1}{2}+\frac{13}{4}\right)=\frac{5}{2}+\frac{1}{11}.\frac{11}{2}+\frac{1}{15}.\frac{15}{4}\)
=> \(B=\frac{5}{2}+\frac{1}{2}+\frac{1}{4}=\frac{10}{4}+\frac{2}{4}+\frac{1}{4}=\frac{13}{4}\)
Ta có
1/7.B = 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7.B = 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7.B = 1/2 - 1/28
1/7.B = 14/28 - 1/28
1/7.B = 13/28
B = 13/28 : 1/7
B = 13/28 . 7
B = 13/4
Ta có: 2003x14 = 2002x14 + 14
=> 2003x14 + 1988 + 2001x2002 = 2002x14 + 2002 + 2001x2002 = 2016x2002
Lại có: 2002+2002x503+504x2002 = 1008x2002
=> 2003x14+1988+2001x2002 2016x2002
-------------------------------- = --------------------- = 2
2002+2002x503+504x2002 1008x2002
B = \(\left(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\right)\))
B = 7 . \(\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
B = 7 . \(\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
B = 7 . ( 1/2 - 1/28 )
B = 7 . 13/28
B = 13/4
1/7B= 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28
1/7B= 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28
1/7B= 1/2 - 1/28= 13/28
B= 13/28 : 1/7= 13/28.7= 13/4
Vậy B= 13/4
Ta có:
\(\frac{1\div2003+1\div2004-1\div2005}{5\div2003+5\div2004-5\div2005}\) - \(\frac{2\div2002+2\div2003-2\div2004}{3\div2002+3\div2003-3\div2004}\)
Đơn giản đi hết ta sẽ còn:
\(\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
2.
Ta có:
Số khoảng cách của các số trong dãy là 23 = 8
=> Tổng của dãy dưới sẽ gấp 8 lần tổng dãy trên.
=> 3025 . 8 = 24200
1/2003.2002 - 1/2002.2001 - ... - 1/3.2 - 1/2.1
= 1/2003.2002 - (1/2002.2001 + ... + 1/3.2 + 1/2.1)
= 1/2002.2003 - (1/1.2 + 1/2.3 + ... + 1/2001.2002)
= 1/2002.2003 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/2001 - 1/2002)
= 1/2002.2003 - (1 - 1/2002)
= 1/2002.2003 - 2001/2002
= 1/2002.2003 - 2001.2003/2002.2003
= 1-2001.2003/2002.2003