Bài này đơn giản thôi mà !
Trong tích các số tự nhiên từ 1 đến 2006 chắc chắn tồn tại 2 thừa số là 223 và 9
mà 2 số này có tích là 223 x 9 = 2007
=> B \(⋮\)2007
Các câu hỏi tương tự
Cho frac{a}{b}frac{c}{d}. Chứng minh:a) frac{left(a-bright)^3}{left(c-dright)^3}frac{3a^2+2b^2}{3c^2+2d^2}b)frac{4a^4+5b^4}{4c^4+5d^4}frac{a^2b^2}{c^2d^2}c)left(frac{a-b}{c-d}right)^{2005}frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}d)frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}frac{left(a+bright)^{2005}}{left(c+dright)^{2005}}e)frac{left(20a^{2006}+11b^{2006}right)^{2007}}{left(20a^{2007}-11b^{2007}right)^{2006}}frac{left(20c^{2006}+11d^{2006}right)^{2007}}{left(20c^{2007}-11d^{2007}right)^{20...
Đọc tiếp
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)Chứng minh:
a)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}^{ }}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
b)\(\left(4a+5b\right)\left(7c-11d\right)=\left(7a-11b\right)\left(4c+5d\right)\)
Tìm x
a/\(\frac{x+7}{2003}+\frac{x+4}{2006}=\frac{x-1}{2011}+\frac{x-5}{2015}\)
b/\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c/\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
So sánh hai số sau:
\(A=\frac{1}{2006}\)
\(B=\frac{1}{2008}+\left(\frac{1}{2008}+\frac{1}{2008^2}\right)^2+...+\left(\frac{1}{2008}+\frac{1}{2008^2}+...+\frac{1}{2008^{2007}}\right)^{2007}\)
So sánh hai số sau:
\(A=\frac{1}{2006}\)
\(B=\frac{1}{2008}+\left(\frac{1}{2008}+\frac{1}{2008^2}\right)^2+...+\left(\frac{1}{2008}+\frac{1}{2008^2}+...+\frac{1}{2008^{2007}}\right)^{2007}\)
1,Tìm số hữu tỉ x biết\(\frac{x+4}{2005}+\frac{x+3}{2006}=\frac{x+2}{2007}+\frac{x+1}{2008}\)
2,tìm x biết:\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
Tình A=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(S=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
1.\(A=\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)
2. \(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
3. Hãy so sánh A và B
\(A=\frac{10^{2006}+1}{10^{2007}+1}\) \(B=\frac{10^{2007}+1}{10^{2008}+2}\)