\(\left(\frac{1}{2x}-1004\right)^{2008}=\left(\frac{1}{2x}-1004\right)^{2006}\)

\(\Rightarrow\left(\frac{1}{2x}-1004\right)^{2008}-\left(\frac{1}{2x}-1004\right)^{2006}=0\)

\(\Rightarrow\left(\frac{1}{2x}-1004\right)^{2006}.\left(\frac{1}{2x}-1004\right)^2-\left(\frac{1}{2x}-1004\right)^{2006}=0\)

\(\Rightarrow\left(\frac{1}{2x}-1004\right)^{2006}.\left[\left(\frac{1}{2x}-1004\right)^2-1\right]=0\)

\(\Rightarrow\left(\frac{1}{2x}-1004\right)^{2006}=0\) hoặc \(\left(\frac{1}{2x}-1004\right)^2-1=0\)

TH1: \(\left(\frac{1}{2x}-1004\right)^{2006}=0\)

\(\Rightarrow\frac{1}{2x}-1004=0\)

\(\Rightarrow\frac{1}{2x}=1004\)

\(\Rightarrow2x=1:1004=\frac{1}{1004}\)

\(\Rightarrow x=\frac{1}{2008}\)

TH2: \(\left(\frac{1}{2x}-1004\right)^2-1=0\)

\(\Rightarrow\left(\frac{1}{2x}-1004\right)^2=1\)

\(\Rightarrow\frac{1}{2x}-1004=1\) hoặc \(\frac{1}{2x}-1004=-1\)

\(\Rightarrow\frac{1}{2x}=1005\) hoặc \(\frac{1}{2x}=1003\)

\(\Rightarrow x=\frac{1}{2010}\) hoặc \(x=\frac{1}{2006}\)

Vậy \(x\in\left\{\frac{1}{2010};\frac{1}{2008};\frac{1}{2006}\right\}\)

Cảm ơn nhìu nhé 

a, 1004 .2009 + 1005 = (1005-1) .2009 +1005

                                = 1005 .2009 -2009 +1005

                                = 1005 .2009 -1004

Vậy ( 1004 .2009 +1005) / (1005 .2009 -1004) =1

b, 1004 .2010 +1 = 1004 .2009 +1004 +1

                          = (1006 -2) .2009 +1005

                          = 1006 .2009 -2 .2009 +1005

                          = 1006 .2009 -4008 +1005

                          = 1006 .2009 -3013

Vậy (1004 .2010 +1) / (1006 .2009 -3013) = 1

c, 2007 .2009 -2 = 2007.(2008+1) -2

                         = 2007.2008 +2007 -2

                         = 2007.2008 +2005

                         = (2008-1) .2008 +2005

                         = 2008 .2008 -2008 +2005

                         = 2008 .2008 -3

Vậy (2008 .2008 -3) / (2007 .2009 -2) =1

bam may đi bạn tui

dựa và mà làm nha bạn

1005.2009-1004=1004.2009+2009-1004=1004.2009+1005

=> a=1

tra loi nhanh mk dang can

\(A=2008.\left(\frac{1}{2007}-\frac{1000}{1004}\right)-2009.\left(\frac{1}{2007}-2\right)\)

\(\Rightarrow A=2008.\frac{1}{2007}-2000-2009.\frac{1}{2007}+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(2008-2009\right)-2000+4018\)

\(\Rightarrow A=\frac{1}{2007}.\left(-1\right)-2000+4018\)

\(\Rightarrow A\approx2018\)

thanh you

\(=\dfrac{2008}{2007}-2\cdot2009-\dfrac{2009}{2007}+2009\cdot2\)

=-1/2007

Áp dụng bđt Cauchy Shwarz dạng Engel, ta có:

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{\left(x^2+y^2\right)^2}{a+b}=\dfrac{1}{a+b}\) (vì \(x^2+y^2=1\))

mà \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) (theo đề bài)

\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) (tính chất của dãy tỉ số bằng nhau)

\(\Rightarrow x^2=\dfrac{a}{a+b}\)

\(B=\dfrac{x^{2008}}{a^{1004}}+\dfrac{y^{2008}}{b^{1004}}\)

\(=\left(\dfrac{x^2}{a}\right)^{1004}+\left(\dfrac{y^2}{b}\right)^{1004}\)

\(=2\times\left(\dfrac{\dfrac{a}{a+b}}{a}\right)^{1004}\) (vì \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\))

Thay số vào ròi tính thoy ~~! (xxx)

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{\left(x^2+y^2\right)^2}{a+b}\left(dox^2+y^2=1\right)\)

\(\Leftrightarrow\dfrac{x^4}{a}+\dfrac{y^4}{b}-\dfrac{\left(x^2+y^2\right)^2}{a+b}=0\)

Tự biến hóa, hô phép ;v

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)

\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{y^{2008}}{b^{2004}}\Rightarrow B=2.\dfrac{x^{2008}}{a^{1004}}\)

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{1}{\left(a+b\right)^{1004}}\)

\(\dfrac{2}{\left(a+b\right)^{1004}}=2.\dfrac{x^{2008}}{a^{1004}}=B\)

Vậy: \(B=\dfrac{2}{\left(a+b\right)^{1004}}\)