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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: \(x+y+z=0\)
\(\Leftrightarrow\) \(\left(x+y+z\right)^2=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2=0\) (vì xy + yz + xz =0)
\(\Leftrightarrow\)\(x=y=z=0\)
Vậy \(S=\left(0-1\right)^{1999}+0^{2003}+\left(0+1\right)^{2006}=0\)
\(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Ta lại có:
\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3=1\)
\(\Leftrightarrow x+y+z=1\)
Làm nốt
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3x^2y-3xy^2-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
Vì \(x+y+z\ne0\) nên \(x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]=0\)
\(\Rightarrow x=y=z\) thay vào P ta được :
\(P=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)
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....
X3 + Y3 + Z3 = 3XYZ
<=> X3 + Y3 + Z3 - 3XYZ = 0
<=> ( X3 + Y3 ) + Z3 - 3XYZ = 0
<=> ( X + Y )3 - 3XY( X + Y ) + Z3 - 3XYZ = 0
<=> [ ( X + Y )3 + Z3 ] - 3XY( X + Y + Z ) = 0
<=> ( X + Y + Z )[ ( X + Y )2 - ( X + Y ).Z + Z2 - 3XY ] = 0
<=> ( X + Y + Z )( X2 + Y2 + Z2 - XY - YZ - XZ ) = 0
<=> \(\orbr{\begin{cases}X+Y+Z=0\\X^2+Y^2+Z^2-XY-YZ-XZ=0\end{cases}}\)
+) X + Y + Z = 0 => \(\hept{\begin{cases}X+Y=-Z\\Y+Z=-X\\X+Z=-Y\end{cases}}\)
KHI ĐÓ : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(\frac{X+Y}{Y}\right)\left(\frac{Y+Z}{Z}\right)\left(\frac{X+Z}{X}\right)=\frac{-Z}{Y}\cdot\frac{-X}{Z}\cdot\frac{-Y}{X}=-1\)
+) X2 + Y2 + Z2 - XY - YZ - XZ = 0
<=> 2( X2 + Y2 + Z2 - XY - YZ - XZ ) = 0
<=> 2X2 + 2Y2 + 2Z2 - 2XY - 2YZ - 2XZ = 0
<=> ( X2 - 2XY + Y2 ) + ( Y2 - 2YZ + Z2 ) + ( X2 - 2XZ + Z2 ) = 0
<=> ( X - Y )2 + ( Y - Z )2 + ( X - Z )2 = 0 (1)
DỄ DÀNG CHỨNG MINH (1) ≥ 0 ∀ X,Y,Z
DẤU "=" XẢY RA <=> X = Y = Z
KHI ĐÓ : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)
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