chung minh : 1/5^3+1/6^3+1/7^3+........+1/2004^3 <1/40
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a) A=21+22+23+...+22010
A=(21+22)+(23+24)+.....+(22009+22010)
A=(21x3)+(23x3)+.....+(22009x3)
A=3x(21+23+.......+22009)
Vậy A chia hết cho 3.
NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !
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a) \(1-2-3+4+5-6-7+...+2001-2002-2003+2004\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2001-2002-2003+2004\right)\)
\(=0+0+...+0=0\)
b) \(1+2-3-4+5+6-7-8+...+2001+2002-2003-2004\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2001+2002-2003-2004\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right)\cdot501=\left(-2004\right)\)
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A = ( 6 : 3/5 - 7/6 * 6/7 ) : ( 21/5 * 10/11 + 57/11 )
A = ( 10 - 1 ) : ( 42/11 + 57/11)
A = 9 : 9
A = 1
B = 59 /10 : 3/2 - ( 7/3 * 9/2 - 2 * 7/3 ) : 7/4
B = 59/15 - ( 21/2 - 14/3 ) : 7/4
B = 59/15 - 35/6 : 7/4
B = 59/15 - 10/3
B = 3/5
Bài toán tổng quát:
Với mọi n\(\in\)N* ta có: \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{n\left(n^2-1\right)}=\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Áp dụng vào bài toán:
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}< \frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
mà \(\frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+...+\frac{1}{2003.2004.2005}\)
\(=\frac{1}{2}\left(\frac{2}{4.5.6}+\frac{2}{5.6.7}+\frac{2}{6.7.8}...+\frac{2}{2003.2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+\frac{1}{6.7}-\frac{1}{7.8}...+\frac{1}{2003.2004}-\frac{1}{2004.2005}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{2003.2004}\right)=\frac{1}{40}-\frac{1}{2.2003.2004}< \frac{1}{40}\)
=>\(\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}+...+\frac{1}{2002.2003.2004}< \frac{1}{40}\)