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\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3x^2y-3xy^2-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
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 :
\(P=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)
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 : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)
Bài 1
\(a^2-2a+6b+b^2=-10\)
<=>\(a^2-2a+1+b^2+6b+9=0\)
<=>\((a-1)^2+(b+3)^2=0\)
Ta lại có: \((a-1)^2\ge0 \)
\((b+3)^2\ge0\)
=> \((a-1)^2+(b+3)^2\ge0\)
Mà\((a-1)^2+(b+3)^2=0\)
=>(a-1)2=0=>a=1
(b+3)2=0=>b=-3
Vậy a=1,b=-3
Bài 2
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
Lời giải:
Ta có:
\(x^3+y^3+z^3=3xyz\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)
Vì \(x+y+z\neq 0\Rightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow 2(x^2+y^2+z^2-xy-yz-xz)=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Ta thấy \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\)
\(\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\). Dấu bằng xảy ra khi
\((x-y)^2=(y-z)^2=(z-x)^2=0\Leftrightarrow x=y=z\)
Khi đó:
\(P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=(1+1)(1+1)(1+1)=8\)