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với xyz=2009, thay vào, ta có
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
=\(\frac{xz}{1+zx+y}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}=1\)
=> ... k phụ thuộc vào x,y,z(ĐPCM)
^_^
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0
\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)
Tương tự rồi cộng lại thì \(BT=0\) thì phải
Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)
Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)
\(=\frac{1}{x^2+\left(y-z\right)\left(y+z\right)}+\frac{1}{y^2+\left(z-x\right)\left(z+x\right)}+\frac{1}{z^2+\left(x-y\right)\left(x+y\right)}\left(4\right)\)
Because \(x^2+y^2+z^2=3xyz\)
\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)ư\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)
The first case: If \(x+y+z=0\left(1\right)\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}\left(2\right)}\)
From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)
\(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have
\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)
\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)
\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)
The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)
We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)
From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)
Because \(x=y=z\Rightarrow x^2=y^2=z^2=xy=yz=zx\)
So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
\(=\frac{z+x+y}{xyz}=0\)
So...
Ta có: x+y+z=0
Suy ra: x+y=-z; y+z=-x; z+x=-y
ta có: \(\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)\left(\frac{z}{x}+1\right)\)\(=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}\)
\(=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}\)
\(=-1\)
Bạn tham khảo tại đây:
Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)
\(\Rightarrow yz+zx+xy=0\)
Ta có : \(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-zx-xy\)
\(=x\left(x-z\right)-y\left(x-z\right)\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự : \(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz\)
\(=y\left(y-x\right)+z\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\)
\(z^2+2xy=\left(x-z\right)\left(y-z\right)\)
\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\) \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
Khai triển cả 2 vế ta được \(\left(\frac{1}{y}+\frac{1}{z}\right)^2+\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)
=>\(\hept{\begin{cases}\frac{1}{y}+\frac{1}{z}=0\\\frac{1}{x}+\frac{1}{z}=0\end{cases}}\)=>\(\frac{1}{x}=\frac{1}{y}\Rightarrow x=y\)
=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{z}=2\Rightarrow\frac{4}{x^2}+\frac{4}{xz}+\frac{1}{z^2}=4\)(1)
\(\frac{2}{xy}-\frac{1}{z^2}=\frac{2}{x^2}-\frac{1}{z^2}=4\)(2)
Từ (1) và (2) suy ra
\(\frac{2}{x^2}+\frac{4}{xz}+\frac{2}{z^2}=0\Rightarrow\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}=0\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)\(\Rightarrow\frac{1}{x}+\frac{1}{z}=0\Rightarrow x=y=-z\)
=> \(P=\left(x+2y+z\right)^{2019}=\left(2y\right)^{2019}\)
à thêm cái này nữa. Sorry viết thiếu
Vì x=y=-z\(\Rightarrow\frac{2}{x}-\frac{1}{x}=2\Rightarrow\frac{1}{x}=2\Rightarrow x=\frac{1}{2}.\)
lúc đó \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{xyc+yza+zxb}{abc}=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\frac{yza+zxb+xyc}{xyz}=0\)
\(\Rightarrow yza+zxb+xyc=0\)
\(\Rightarrow A=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)