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\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+...+\dfrac{x-2016}{1}=2016\)
\(\Leftrightarrow\dfrac{x-1}{2016}-1+\dfrac{x-2}{2015}-1+\dfrac{x-3}{2014}-1+...+\dfrac{x-2016}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}+...+\dfrac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow x-2017=0\) (do \(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\ne0\))
\(\Rightarrow x=2017\)
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2016}\)
\(\Rightarrow\dfrac{bc+ac+bc}{abc}=\dfrac{1}{2016}\)
\(\Rightarrow\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)
\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)
\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)
-Vậy trong ba số a,b,c tồn tại 2 số đối nhau.
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1
\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2015x+2017}{x^2-4}\)
\(B=B_1+B_2+...+B_{2016}\)
\(B_1=\dfrac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}\)
\(B_1=\sqrt{x+1}-\sqrt{x}\)
\(B_2=\sqrt{x+2}-\sqrt{x+1}\)
\(B_3=\sqrt{x+3}-\sqrt{x+2}\)
...
\(B_{2015}=\sqrt{x+2015}-\sqrt{x+2014}\)
\(B_{2016}=\sqrt{x+2016}-\sqrt{x+2015}\)
\(B=\sqrt{x+2016}-\sqrt{x}\)
\(B\left(2017\right)=\sqrt{2017+2016}-\sqrt{2017}\)
Lời giải:
Ta thấy:
\(\frac{1}{2016^x+1}+\frac{1}{2016^{-x}+1}=\frac{1}{2016^x+1}+\frac{1}{\frac{1}{2016^x}+1}=\frac{1}{2016^x+1}+\frac{2016^x}{1+2016^x}=\frac{2016^x+1}{2016^x+1}=1\)
Do đó:
\(A=\frac{1}{2016^{-2016}+1}+\frac{1}{2016^{-2015}+1}+...+\frac{1}{2016^{-1}+1}+\frac{1}{2016^0+1}+\frac{1}{2016^1+1}+...+\frac{1}{2016^{2016}+1}\)
\(=\underbrace{\left(\frac{1}{2016^{-2016}+1}+\frac{1}{2016^{2016}+1}\right)+\left(\frac{1}{2016^{-2015}+1}+\frac{1}{2016^{2015}+1}\right)+....+\left(\frac{1}{2016^{-1}+1}+\frac{1}{2016^{1}+1}\right)}_{ \text{2016 cặp}}+\frac{1}{2016^0+1}\)
\(=1.2016+\frac{1}{1+1}=2016+\frac{1}{2}=\frac{4033}{2}\)
Em cảm ơn nhiều ạ