Nếu f(x)=x+2x2-3x3-4x4+5x5+6x6-7x7-8x8+....+2009x2009+2010x2010-2011x2011-2012x2012 chia cho g(x)=x-1 thì dư bao nhiêu?
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1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 385
1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10
= 1+4+9+16+25+36+49+64+81+100
=(81+9)+(64+16)+(49+1)+)36+4)+25+100
=90+80+50+40+25 +100
=385
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![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(A
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1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
\(1x1=1\)
\(2x2=4\)
\(3x3=9\)
\(4x4=16\)
\(5x5=25\)
\(6x6=36\)
\(7x7=49\)
\(8x8=64\)
\(9x9=81\)
\(10x10=100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy:
1/2*2<1/1*2)vì 2*2>1*2).
1/3*3<1/2*3(vì 3*3>2*3).
...
1/8*8<1/7*8(vì 8*8>7*8).
=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.
=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.
=>B<1-1/8.
=>B<7/8.
Mà 7/8<1.
=>B<1.
Vậy B<1(đpcm).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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1.
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{2012}{2013}\)
\(A=\frac{1.2.3.4.....2012}{2.3.4.5......2013}\)
\(A=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(B=\frac{2012\left(2013-2012\right)}{2012\left(2011+2\right)}\)
\(B=\frac{2013-2012}{2011+2}\)
\(B=\frac{1}{2013}\)
\(Vì:\frac{ 1}{2013}=\frac{1}{2013}\)
\(\Rightarrow\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(Hay: A=B\)
\(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\frac{4\times5}{5\times5}\times...\times\frac{2012\times2013}{2013\times2013}\)
\(\Rightarrow A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2012}{2013}\)
\(\Rightarrow A=\frac{1\times2\times3\times4\times...\times2012}{2\times3\times4\times5\times...\times2013}\)
\(\Rightarrow A=\frac{1}{2013}\)
\(B=\frac{2012\times2013-2012\times2012}{2012\times2011+2012\times2}\)
\(\Rightarrow B=\frac{2012\times\left(2013-2012\right)}{2012\times\left(2011+2\right)}\)
\(\Rightarrow B=\frac{2012\times1}{2012\times2013}\)
\(\Rightarrow B=\frac{1}{2013}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng định lý Bezout, số dư của phép chia f(x) cho g(x) là \(f\left(1\right)\)
\(f\left(1\right)=1+2-3-4+...-2011-2012\)
\(=-2-2-2-....-2\) (\(\frac{2012}{2}=1006\) số -2)
\(=-2012\)
Vậy số dư là \(-2012\)