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Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))
Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) .
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0
Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
Thay x+y+z=2020 vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2020}\) có:
\(\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)
<=>\(\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)
<=>\(x^2y+xy^2+xyz+xyz+y^2z+yz^2+x^2z+xyz+xz^2=xyz\)
<=>\(xy\left(x+y\right)+z^2\left(x+y\right)+y^2z+x^2z+3xyz-xyz=0\)
<=>\(\left(x+y\right)\left(xy+z^2\right)+z\left(y^2+x^2+2xy\right)=0\)
<=>\(\left(x+y\right)\left(xy+z^2\right)+z\left(x+y\right)^2=0\)
<=>\(\left(x+y\right)\left(xy+z^2+xz+yz\right)=0\)
<=>\(\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=0\)
<=>\(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
=> \(\left[{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)
Tại x=-y => \(x^{2009}=-y^{2009}\)
<=>\(x^{2009}+y^{2009}\)=0
Có \(P=\left(x^{2009}+y^{2009}\right)\left(y^{2011}+z^{2011}\right)\left(z^{2013}+x^{2013}\right)=0\left(y^{2011}+z^{2011}\right)\left(z^{2013}+x^{2013}\right)=0\)
Tương tự các trường hợp kia cũng => P=0
Vậy P=0
\(A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
=> \(\left(-A\right)=\frac{yz}{\left(x-y\right)\left(z-x\right)}+\frac{xz}{\left(x-y\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(y-z\right)}\)
<=> \(\left(-A\right)=\frac{yz\left(y-z\right)+xz\left(z-x\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{y^2z-yz^2+xz^2-x^2z+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
<=> \(\left(-A\right)=\frac{z^2\left(x-y\right)-z\left(x^2-y^2\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)=> \(\left(-A\right)=\frac{\left(x-y\right)\left(z^2-zx-zy+xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left[z\left(z-x\right)-y\left(z-x\right)\right]}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(\left(-A\right)=\frac{\left(x-y\right)\left(z-x\right)\left(z-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=-1\)
=> A = 1
Đáp số: A=1
