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\(\Leftrightarrow\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)
\(\Leftrightarrow\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x=2012\)
Có điều kiện là a>0 và b>0 nữa nha
Theo bđt cô si ta có : \(a+b\ge2\sqrt{ab}\) (1)
\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\) (2)
Nhân vế theo vế 1 và 2 ta có : \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2\sqrt{ab}\cdot2\sqrt{\frac{1}{ab}}=4\cdot\sqrt{\frac{ab}{ab}}=4\)
Vậy \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\) đpcm
ta có\(\frac{x-2013}{-3}+\frac{x-2012}{-4}=\frac{x-2011}{-5}-\frac{x-1}{-2015}\)
\(\Leftrightarrow\frac{x-2013}{-3}+1+\frac{x-2012}{-4}+1=\frac{x-2011}{-5}+1-\frac{x-1}{-2015}+1\)
\(\Leftrightarrow\frac{x-2013-3}{-3}+\frac{x-2012-4}{-4}=\frac{x-1-2015}{-5}-\frac{x-1-2015}{-2015}\)
\(\Leftrightarrow\frac{x-2016}{-3}+\frac{x-2016}{-4}=\frac{x-2016}{-5}-\frac{x-2016}{-2015}\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{-3}+\frac{1}{-4}-\frac{1}{-5}+\frac{1}{-2015}\right)=0\)
\(\Leftrightarrow x-2016=0\)
\(\Leftrightarrow x=2016\)
Vậy tập nghiệm của phương trình đã cho là là:\(S=\left(2016\right)\)
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
a) \(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)=-3
⇔\(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)+3=0
⇔\(\frac{x+2}{2002}\)+1+\(\frac{x+5}{1999}\)+1+\(\frac{x+201}{1803}\)+1=0
⇔\(\frac{x+2004}{2002}\)+\(\frac{x+2004}{1999}\)+\(\frac{x+2004}{1803}\)=0
⇔(x+2004)(\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))=0
Mà (\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))≠0
⇒x+2004=0
⇔x=-2004
Vậy tập nghiệm của phương trình đã cho là:S={-2004}
Ta có:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+..+\frac{2}{2013}+\frac{1}{2014}\)
\(=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{2}{2013}+1\right)+\left(\frac{1}{2014}+1\right)+1\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)\)
Do đó: \(A=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}}=2015\)
a.D=4a(3+b)+a*2a-3ab=12a+4ab+2a2-3ab=2a2+ab+12a=a(2a+b+12)
b.bạn viết đề kiểu j vậy
Ko sai đề nha bn