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ES
Cho f( x ) = x mũ 2005- 2006.x mũ 2004+ 2006.x mũ 2003-....- 2006.x mũ 2+ 2006.x mũ 1.
Tính f( 2005)
1
21 tháng 6 2022
x=2005
nên x+1=2006
\(f\left(x\right)=x^{2005}-x^{2004}\left(x+1\right)+x^3\left(x+1\right)-...+x\left(x+1\right)\)
\(=x^{2005}-x^{2005}-x^{2004}+x^{2004}+...-x^3-x^2+x^2+x\)
=x=2005
KN
24 tháng 6 2020
Ta có :
\(x=2005\Rightarrow x+1=2006\)
Thay \(2006=x+1\) vào biểu thức trên ta được :
\(x^{2005}-\left(x+1\right)x^{2004}+\left(x+1\right)x^{2003}-\left(x+1\right)x^{2002}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2005}-x^{2005}+x^{2004}-x^{2004}+x^{2003}-...-x^3+x^2-x^2+x-1\)
\(=x-1\) mà \(x=2005\)
\(\Rightarrow x^{2005}-2006.x^{2004}+2006.x^{2003}-2006.x^{2002}+...-2006.x^2+2006x-1=2005-1=2004\)
10 tháng 12 2015
Thay x=2005 vào biểu thức, ta được:
20052005-2006*20052004+...+2006*20052-2006*2005-1
=20052005-(2006*20052004-..-2006*20052+2006*2005+1)
Đặt A=(2006*20052004-..-2006*20052+2006*2005+1)
2005A=2006*20052005-..-2006*20053+2006*20052+2005
2005A+2005*2006=2006*20052005-..-2006*20053+2006*20052+2006*2005+1+2004=A+2004
2005A-A=2004-2005*2006
2004A=2004-2005*2006
A=(2004-2005*2006)/2004=1-(2005*2006)/2004
=>20052005-(2006*20052004-..-2006*20052+2006*2005+1)=20052005-1+(2005*2006)/2004
đến đây cậu làm được chưa, quy đồng lên rồi tính, phân phối ra ý
5 tháng 8 2017
\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)
\(\Rightarrow x-2009=0\Rightarrow x=2009\)
6 tháng 8 2017
\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)
\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)
\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2009=0\)
\(\Leftrightarrow x=2009\)
Vậy \(x=2009\)
10 tháng 12 2015
\(A=x^{2005}-2005x^{2004}-x^{2004}+2005x^{2003}+x^{2003}-2005x^{2002}-.....+x^3-2005x^2-x^2+2005x+x-2005+2004\)\(=\left(x-2005\right)x^{2004}-\left(x-2005\right)x^{2003}+\left(x-2005\right)x^{2002}-....+\left(x-2005\right)x^2-\left(x-2005\right)x+\left(x-2005\right)+2004\)\(=\left(x-2005\right)\left(x^{2004}-x^{2003}+x^{2002}-......+x^2-x+1\right)+2004\)
Với x = 2005 => x - 2005 =0
=> A =2004
25 tháng 3 2018
Ta có :
\(\frac{x+3}{2003}+\frac{x+2}{2004}+\frac{x+1}{2005}=-3\)
\(\Leftrightarrow\)\(\left(\frac{x+3}{2003}+1\right)\left(\frac{x+2}{2004}+1\right)\left(\frac{x+1}{2005}+1\right)=-3+3\)
\(\Leftrightarrow\)\(\frac{x+2006}{2003}+\frac{x+2006}{2004}+\frac{x+2006}{2005}=0\)
\(\Leftrightarrow\)\(\left(x+2006\right)\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)=0\)
Vì \(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\ne0\)
Nên \(x+2006=0\)
\(\Rightarrow\)\(x=-2006\)
Vậy \(x=-2006\)
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