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1/x + 1/y + 1/z = 1/x+y+z
<=> xy+yz+zx/xyz = 1/x+y+z
<=> (xy+yz+xz).(x+y+z)=xyz
<=> x^2y+xy^2+y^2z+z^2y+z^2x+x^2z+3xyz=xyz
<=> x^2y+y^2x+y^2z+z^2y+z^2x+x^2z+2xyz = 0
<=> (x+y).(y+z).(z+x) = 0
<=> x+y=0 hoặc y+z=0 hoặc x+z=0
<=> x=-y hoặc y=-z hoặc z=-x
Nếu x=-y => x^25 = -y^25 => P = 0
Nếu y=-z => y^3 = -z^3 => P = 0
Nếu z=-x => z^2006 = x^2006 => P = 0
Vậy P = 0
Tk mk nha
Ta có:
\(xy+yz+zx=\frac{\left(x+y+z\right)^2-x^2-y^2-z^2}{2}=\frac{7^2-23}{2}=13\)
Ta lại có:
\(xy+z-6=xy+z+1-x-y-z=\left(x-1\right)\left(y-1\right)\)
\(\Rightarrow A=\frac{1}{\left(x-1\right)\left(y-1\right)}+\frac{1}{\left(y-1\right)\left(z-1\right)}+\frac{1}{\left(z-1\right)\left(x-1\right)}\)
\(=\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}=-1\)
Ta có: \(\left(x+y+z\right)=a\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\)
\(\Leftrightarrow\left(xy+yz+zx\right)=\frac{a^2-\left(x^2+y^2+z^2\right)}{2}=\frac{a^2-b^2}{2}\)
Ta lại có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\)
\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{c}\)
\(\Leftrightarrow xyz=c\left(xy+yz+zx\right)=c.\frac{a^2-b^2}{2}\)
Ta biến đổi: \(x^3+y^3+z^3=x^3+y^3+z^3-3xyz+3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+zx\right)\right)+3xyz\)
\(=a.\left(b^2-\frac{a^2-b^2}{2}\right)+\frac{3c\left(a^2-b^2\right)}{2}\)
\(\left\{\begin{matrix}x+y+z=a\left(1\right)\\x^2+y^2+z^2=b^2\left(2\right)\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\left(3\right)\end{matrix}\right.\)
HĐT ta có\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+xz+yz\right)\right)-3xyz\)
từ (3)=> (xy+xz+yz)/(xyz)=1/c(*)
(1) bình phường=>2(xy+xz+yz)=(a^2-b^2 )
(*)=> xyz=(a^2-b^2).c/2
Thay hết vào biểu thức trên => đáp số
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)=1\)
\(\Leftrightarrow3xyz+yz\left(y+z\right)+xz\left(x+z\right)+xy\left(x+y\right)=xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) hay B = 0
Đặt \(\left\{{}\begin{matrix}xy=a\\yz=b\\zx=c\end{matrix}\right.\)
Giả thiết \(\Leftrightarrow a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+3a^2b+3ab^2+c^3-3abc-3a^2b-3ab^2=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
+) TH1: \(a+b+c=0\Leftrightarrow xy+yz+zx=0\)
Biến đổi linh tinh P chắc là ra :D
+) TH2: \(a=b=c\Leftrightarrow xy=yz=zx\Leftrightarrow x=y=z\)
\(P=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{z+x}{x}=\frac{2y}{y}\cdot\frac{2z}{z}\cdot\frac{2x}{x}=2\cdot2\cdot2=8\)
Vậy....
TH1: \(xy+yz+zx=0\)
\(\Leftrightarrow z\left(x+y\right)=-xy\)
\(\Leftrightarrow x+y=\frac{-xy}{z}\)
Vì vai trò của x, y, z là như nhau nên ta cũng có :
\(\left\{{}\begin{matrix}y+z=\frac{-yz}{x}\\z+x=\frac{-zx}{y}\end{matrix}\right.\)
Ta có \(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(P=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{z+x}{x}\)
\(P=\frac{\frac{-xy}{z}\cdot\frac{-yz}{x}\cdot\frac{-zx}{y}}{xyz}\)
\(P=\frac{\frac{-x^2y^2z^2}{xyz}}{xyz}\)
\(P=\frac{-xyz}{xyz}=-1\)
Vậy....