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\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)
\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)
\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)
\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(M=11\cdot\left(1-\frac{1}{103}\right)\)
\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3.\frac{63}{268}\)
\(\Leftrightarrow Q=\frac{189}{268}\)
Câu b) bạn làm tương tự nhé :)
A=3²/1.4+3²/4.7+3²/7.10+...+3²/97.100
A=9/1.4+9/4.7+9/7.10+...+9/97.100
A=9x(1/1.4+1/4.7+1/7.10+...+1/97.100)
A=9x(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
A=9x(1-1/100)
A=9x99/100
A=9x33/100
A=297/10=2,97
(3^2)/1.4+(3^2)/4.7+...+(3^2)/97.100
=3.(3/4.7+3/7.10+...+3/97.100)
=3.(1/4-1/7+1/7-1/10+...+1/97-1/100)
=3.(1/4-1/100)
=3.6/25=18/25
= 3/1 - 3/4 + 3/4 - 3/7 + 3/7 - 3/10 + ... + 3/97 - 3/100 = 3/1 - 3/100 = 297/100
\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)=\frac{297}{100}\)
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+..+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)=3.\dfrac{99}{100}=\dfrac{297}{100}\)
Giải
Ta gọi T = (1^2+2^2+...+2005^2)-(1.3+2.4+3.5+...+2004.2006)
Đặt A = 1^2+2^2+3^2+...+2005^2
=> A = 1.1 + 2.2 +3.3 +...+ 2005.2005
=> A = 1.(2-1) + 2.(3-1) + 3.(4-1) +...+ 2005.(2006-1)
==> A = 1.2-1.1 + 2.3-1.2 + 3.4-1.3+...+2005.2006-1.2005
=> A = (1.2+2.3+3.4+...+2005.2006)-(1+2+3+...+2005)
Xét 1.2 +2.3+3.4+...+2005.2006
= 1/3.(1.2.3+2.3.3+...+2005.2006.3)
=1/3.[1.2.(3-0)+2.3.(4-1)+...+2005.2006.(2007-2004)]
=1/3.(1.2.3+2.3.4-1.2.3+...+2005.2006.2007-2004.2005.2006)
= 1/3 . 2005.2006.2007
= 2005.2006.2007/3 = 2690738070
Vậy A= 2690738070 - (1+3+5+...+2005)
=> A= 2690738070- [(2005-1):2+1].(2005+1)/2
=> A = 2690738070 - 1006009
=> A = 2689732061
Đắt B = 1.3+2.4+3.5+4.6+...+2003.2005 +2004.2006
=> B= (1.3+3.5+...+2003.2005)+(2.4+4.6+...+2004.2006)
=> 6B = (1.3.6+3.5.6+...+2003.2005.6)+(2.4.6+4.6.6+...+2004.2006.6)
=> 6B = [1.3.(5+1)+3.5.(7-1)+...+2003.2005.(2007-2001)] + [2.4.(6-0)+4.6.(8-2)+...+2004.2006.(2008-2002)]
=> 6B = (1.3.5+1.3.1+3.5.7-1.3.5+...+2003.2005.2007-2001.2003.2005)+(2.4.6+4.6.8-2.4.6+...+2004.2006.2008-2002.2004.2006)
=> 6B = 1.3.1+2003.2005.2007 + 2004.2006.2008
=> 6B = 16132350300
=> B = 16132350300/6 = 2688725050
Vì T = A - B = 2689732061-2688725050
=> T = 1007011
Q=3/4.7+3/7.10+...+3/64+67
Q=1/4-1/7+1/7-1/10+.....+1/64-1/67
Q=1/4-1/67
Q=63/268
M=22/1.3+22/3.5+..+22/101.103
M=2.(2/1.3+2/3.5+...+2/101+103)
M=2.(1-1/3+1/3-1/5+1/5-......+1/101-1/103)
M=2.(1-1/103)
M=2.102/103
M=204/103
K MÌNH NHA