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 : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)
Khi x + y + z = 0
=> x + y = -z
=> x + z = - y
=> y + z = - x
Khi đó M = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)
ĐỀ THẾ MÀ BẢO m = 8
x3 + y3 + z3 = 3xyz
=> (x + y + z)(x2 - y2 - z2 - xy - yz - zx) = 0
mà x + y + z = 0
=> sao lại x2 - y2 - z2 - xy - yz - zx = 0 được
=> Không có x = y = z
=> M \(\ne\)8
ĐỀ BÀI LÀ X + Y + Z ≠ 0 À :V NHỚ VIẾT RÕ RA NHỚ
BẠN BỎ PHẦN X + Y + Z = 0 GIÙM MÌNH NHÉ
bạn quỳnh sephera có thể giải thích cho minh từ dòng thứ 3 xuống dòng thứ 4 kĩ hơn đc ko
đề là x+y+z=0 bạn ơi
Ta có:\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow x^3+y^3+z^3+3x^2y-3x^2y+3xy^2-3xy^2-3xyz=0\)
\(\Leftrightarrow\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
Vì \(x+y+z=0\)\(\Rightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]=0\)
\(\Rightarrow x=y=z\)Thay vào \(M\)ta được:
\(M=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)
Từ \(x^3+y^3+z^3=3xyz\)
\(\Rightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)^3-3\left(x+y\right).z\left(x+y+z\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(x+y\right)z-3xy\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3yz-3xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
TH1: Nếu \(x+y+z=0\)\(\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}}\)
\(\Rightarrow M=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\)
\(=\frac{\left(-x\right).\left(-y\right).\left(-z\right)}{xyz}=\frac{-xyz}{xyz}=-1\)
TH2: Nếu \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)\(\Rightarrow x=y=z\)
\(\Rightarrow M=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{xyz}\)
\(=\frac{8xyz}{xyz}=8\)
Vậy \(M=1\)hoặc \(M=8\)
