(x-1)(2x²-3x+2016) =0

x-1 =0 => x = 1

còn lại 2x² -3x +2016 =0

giải theo denta cho nhanh

Ta có

(x -1)^2016 >0; (2y-1)^2016>0;  /x+2y-z/^2017>0

Mà tổng ba số trên bằng 0

=>(x-1)^2016=0 ; (2y-1)^2016=0; /x+2y-z/=0

=>x=1; y=1/2; z= 2

\(\Leftrightarrow\left(x-1\right)\left(2x^2-3x+2013\right)=0\Leftrightarrow\left(x-1\right)\left(2\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}\right)+2013-\frac{9}{8}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2\left(x-\frac{3}{4}\right)^2+2011\frac{7}{8}\right)=0\)(1)

Do \(\left(2\left(x-\frac{3}{4}\right)^2+2011\frac{7}{8}\right)>0\forall x\)nên

(1) <=> x - 1 = 0 <=> x = 1.

Ta có \(x=\dfrac{2016}{x\times\left(x+1\right)\times\left(x+2\right)\times........\times\left(x+2016\right)}\)

\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times...........\times2016}\)

Vì x > 0=> \(\left(x+1\right)\times\left(x+2\right)\times...\times\left(x+2016\right)>1\times2\times...\times2016\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{1}{1\times2\times..........\times2016}\)\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{2016}{1\times2\times......\times2016}\)

\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)

Ta có \(x=\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times....\times\left(x+2016\right)}\)

\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times.....\times2016}\)

Vì x>0=>(x+1)×(x+2)×.............×(x+2016) >\(1\times2\times.....\times2016\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{1}{1\times2\times......\times2016}\)

\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{2016}{1\times2\times......\times2016}\)

\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)

\(\left\{{}\begin{matrix}\left(x-1\right)^{2016}\ge0\\\left(2y-1\right)^{2016}\ge0\\\left|x+2y-z\right|^{2017}\ge0\end{matrix}\right.\Rightarrow\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}\ge0\)

Mà \(\left(x-1\right)^{2017}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^{2016}=0\\\left(2y-1\right)^{2016}=0\\\left|x+2y-z\right|^{2017}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\\z=2\end{matrix}\right.\)

\(\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}=0\)

\(\left\{{}\begin{matrix}\left(x-1\right)^{2016}\ge0\\\left(2y-1\right)^{2016}\ge0\\\left|x+2y-z\right|^{2017}\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-1\right)^{2016}=0\\\left(2y-1\right)^{2016}=0\\\left|x+2y-z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\\z=\dfrac{3}{2}\end{matrix}\right.\)

Vậy...

⑴⑵⑶⑷⑸⑹⑺⑻⑼0-

Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)

Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)

\(\Rightarrow xy+yz+zx=2016\)thay vào :

\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)

\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0

Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)

\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)

Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)