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Lời giải:
Từ điều kiện đề bài suy ra:
\(\left\{\begin{matrix} x^{2016}+y^{2016}-x^{2017}-y^{2017}=0\\ x^{2017}+y^{2017}-x^{2018}-y^{2018}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x^{2016}(1-x)+y^{2016}(1-y)=0\\ x^{2017}(1-x)+y^{2017}(1-y)=0\end{matrix}\right.\)
\(\Rightarrow x^{2016}(1-x)(1-x)+y^{2016}(1-y)(1-y)=0\) (trử theo vế)
\(\Leftrightarrow x^{2016}(1-x)^2+y^{2016}(1-y)^2=0\)
Dễ thấy \(x^{2016}(1-x)^2; y^{2016}(1-y)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:
\(x^{2016}(1-x)^2=y^{2016}(1-y)^2=0\)
\(\Rightarrow (x,y)=(0,1), (0,0), (1,1)\) và hoán vị của nó
Thử lại vào đk ban đầu thấy thỏa mãn
Do đó: \(A=x^{2019}+y^{2019}\in\left\{0; 1;2\right\}\)
Vì \(x^{2016}+y^{2016}=x^{2017}+y^{2017}=x^{2018}+y^{2018}\left(x,y\ge0\right)\)
\(\Rightarrow x=y=1\)
\(\Rightarrow A=1^{2019}+1^{2019}\)
\(\Rightarrow A=2\)
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
Ta có \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)
<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)
<=>x\(\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Vì \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\) không thể bằng 0
Vậy x=0
Ta có 1 nghiệm thỏa mãn S=\(\left\{0\right\}\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
vì 1/2015+1/2016-1/2017-1/2018 khác 0
=>x-2014=0<=>x=2014
vậy.....................
chúc bạn học totts ^^
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)
\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy........
Ta có: \(x+2y+3x=0\Leftrightarrow x=-\left(2y+3z\right)\)
Lại có: \(2xy+6yz+3xz=0\Leftrightarrow x\left(2y+3z\right)+6yz=0\)
\(\Leftrightarrow-\left(2y+3z\right)\left(2y+3z\right)+6yz=0\Leftrightarrow-\left(2y+3z\right)^2+6yz=0\)
\(\Leftrightarrow\left(2y+3z\right)^2-6yz=0\Leftrightarrow4y^2+12yz+9z^2-6yz=0\)
\(\Leftrightarrow4y^2+6yz+9z^2=0\Leftrightarrow\left(2y+\dfrac{3z}{2}\right)^2+\dfrac{27z^2}{4}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+\dfrac{3z}{2}\right)^2=0\\\dfrac{27z^2}{4}=0\end{matrix}\right.\) \(\Rightarrow y=z=0\Rightarrow x=0\)
\(\Rightarrow S=\dfrac{\left(-1\right)^{2019}-1^{2017}+\left(-1\right)^{2015}}{1^{2018}+2.0^{2016}+0^{2014}+2}=\dfrac{-1-1+-1}{1+0+0+2}=\dfrac{-3}{3}=-1\)
Vì \(x^{2015}+y^{2015}=x^{2016}+y^{2016}=x^{2017}+y^{2017}\)
\(\Rightarrow x=y=1\) hoặc \(x=y=0\)
Với \(x=y=1\)
\(S=2018\left(1^{2018}+1^{2018}\right)\)
\(S=2018.2\)
\(S=4036\)
Với \(x=y=0\)
\(S=2018\left(0^{2018}+0^{2018}\right)\)
\(S=0\)