\(\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...........+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.............+\frac{1}{2006}}\)
Những câu hỏi liên quan
Tinh A = \(\frac{\frac{2006}{1}+\frac{2006}{2}+\frac{2006}{3}+........\frac{2006}{2006}+\frac{2006}{2007}}{\frac{1}{2006}+\frac{2}{2005}+\frac{3}{2004}+.........+\frac{2005}{2}+\frac{2006}{1}}\)
tính hợp lí C=\(\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+...+\frac{1}{2006}}\)
\(C=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}}\)
Đặt N = \(\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}\)
\(\Rightarrow N=\frac{1}{2006}+.....+\frac{2004}{3}+\frac{2005}{2}+\frac{2006}{1}\)
\(\Rightarrow N=\left(\frac{1}{2006}+1\right)+.....+\left(\frac{2004}{3}+1\right)+\left(\frac{2005}{2}+1\right)+1\)( Có 2005 nhóm )
\(=\frac{2007}{2006}+....+\frac{2007}{3}+\frac{2007}{2}+\frac{2007}{2007}\)
\(=2007\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2006}+\frac{1}{2007}\right)\)
Đặt M = \(\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}\)
\(=2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)\)
Thay N và M vào C , ta có :
\(C=\frac{N}{M}=\frac{2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)}{2007\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}\right)}=\frac{2006}{2007}\)
\(\Rightarrow C=\frac{2006}{2007}\)
Vậy : \(C=\frac{2006}{2007}\)
Tinh A = \(\frac{\frac{2006}{1}+\frac{2006}{2}+.......+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.......\frac{1}{2006}}\)
C=\(\frac{\frac{2006}{2}}{\frac{2006}{1}}\) +\(\frac{2006}{\frac{3}{\frac{2005}{2}}}\) +\(\frac{2006}{\frac{4}{\frac{2004}{3}}}\) +...+\(\frac{2006}{\frac{2007}{\frac{1}{2006}}}\)
Tính :
C=\(\frac{\frac{2006}{3}+\frac{2006}{3}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+...+\frac{1}{2006}}\)
Ta có :
Tử số = \(\frac{2006}{2}+...+\frac{2006}{2007}\)
= 2006.(\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\))
MS= \(\frac{2006}{1}+\frac{2005}{2}+...+\frac{1}{2006}\)
= 2006+\(\frac{2007-2}{2}+\frac{2007-3}{3}+...+\frac{2007-2006}{2006}\)
=200+.(\(\frac{2007}{2}+\frac{2007}{3}+...+\frac{2007}{2006}\)) - ( 1+1+1+...+1 )(2006c/s1)
= 2006 . (\(\frac{2007}{2}+...+\frac{2007}{2006}\))-2006
=\(\frac{2007}{2}+...+\frac{2007}{2006}\)
=2007.(\(\frac{1}{2}+...+\frac{1}{2006}\))
Khi đó :
C= .... bạn tự đáp số
và cuối cùng C = \(\frac{2006}{2007}\)
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Tính :
\(\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2006}{2}+\frac{2006}{3}+...+\frac{1}{2006}}\)
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)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
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Tính một cách hợp lí giá trị của các biểu thức sau:
A=3+6+9+12+...+2007
B=2.53.12+4.6.87-3.8.40
C=(\(\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2006}{2}+\frac{2006}{3}+...+\frac{1}{2006}}\)
A = 3 + 6 + 9 + ... + 2007
=>A = 3( 1 + 2 + 3 + ... + 669 )
=> A = \(3\cdot\left(\frac{670\cdot669}{2}\right)\)
=> A = \(3\cdot224115\)= 672345
B = \(2\cdot53\cdot12+4\cdot6\cdot87-3\cdot8\cdot40\)
=> B = 24 * 53 + 24 * 87 - 24 * 40
=> B = 24 * ( 53 + 87 - 40 )
=> B = 24 * 100 = 2400
c) ta có Tử số = \(2006\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)\)
Mẫu số = \(\frac{2007-1}{1}\)+\(\frac{2007-2}{2}\)+...+\(\frac{2007-2006}{2006}\)
=> Mẫu số = \(\frac{2007}{1}\)\(-1\)+ \(\frac{2007}{2}\)\(-1\)+ ... + \(\frac{2007}{2006}\)\(-1\)
=> Mẫu số = \(\frac{2007}{1}\)+ \(\frac{2007}{2}\)+ ... + \(\frac{2007}{2006}\)- ( 1 + 1 + 1 + ... + 1 ) ( 1 + 1 + ... + 1 có 2006 số hạng 1 )
=> Mẫu số = ( 2007 - 2006 ) + \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)
=> Mẫu số = \(\frac{2007}{2007}\)+ \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)
=> Mẫu số = \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)\)
=> C = \(\frac{TS}{MS}\)= \(\frac{2006}{2007}\)
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Giải phương trình sau :
\(\frac{x^2-2008}{2007}+\:\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\:\frac{x^2-\:2005}{2004}+\:\frac{x^2-2004}{2003}+\:\frac{x^2-2003}{2002}\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .