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\(x^3+y^3+z^3=3xyz\)
⇔ \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz=0\)
⇔ \(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2-2xy+y^2+z^2-2xz+x^2+y^2-2yz+z^2\right)=0\) ⇔ \(\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Do : x , y , z là ba số thực phân biệt , ta có :
\(x+y+z=0\)
⇔ \(x+y=-z;y+z=-x;x+z=-y\)
Khi đó , ta có : \(P=\dfrac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\dfrac{2016xyz}{-xyz}=-2016\)
Ta có: \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)^3-3.\left(x+y\right).z.\left(x+y+z\right)-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right).\left[\left(x+y+z\right)^2-3.\left(x+y\right).z\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2zx-3xz-3yz-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2-xz-yz-xy\right)=0\)
+ \(x+y+z=0\)\(\Rightarrow\)\(C=\frac{x^{2019}+y^{2019}+z^{2019}}{0}\)( Loại )
+ \(x^2+y^2+z^2-xz-yz-xy=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xz-2yz-2xy=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\)\(x=y=z\)
\(\Rightarrow\)\(C=\frac{x^{2019}+x^{2019}+x^{2019}}{\left(x+x+x\right)^{2019}}=\frac{3.x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
Vậy.......
Từ x3 + y3 + z3 = 3xyz
=> ( x + y + z )( x2 + y2 + z2 - xy - yz - xz ) = 0 ( phân tích như bạn kia )
Vì x + y + z ≠ 0
=> x2 + y2 + z2 - xy - yz - xz = 0
<=> 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz = 0
<=> ( x - y )2 + ( y - z )2 + ( x - z )2 = 0
VT ≥ 0 ∀ x,y,z. Đẳng thức xảy ra <=> x=y=z
Khi đó \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)
Đề chưa chuẩn: tuy nhiên đánh vào -2016 => đáp án đúng:
Vì bản chất như sau:
thỏa ĐK ban đầu x^3+y^3+z^3=3xzy
Từ HĐT=>
\(\orbr{\begin{cases}x+y+z=0\left(1\right)\\x^2+y^2+z^2-xy-yz-xz=0\left(2\right)\end{cases}}\)
=>(1)&(2) đều có cặp nghiệm x=y=z=0 khi đó P không xác định
do vậy đề thiếu điều kiện x,y,z không đồng thời =0:(*)
Nếu thêm đk (*) giải tiếp
(2) vô nghiệm
do vậy khi đó chỉ có nghiệm duy nhất của (1)
x+y=-z
x+z=-y
z+y=-x
Thay vào biểu thwucs P=-2016
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
\(\Rightarrow A=0\)
(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0
(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0
(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0
(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0
Suy ra x+y+z =0
x+y = -z
y+z = -x
x+z = -y
B = -16 + (-3) +2038 = 2019
Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)
+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)
\(=-16-3+2038=2019\)
+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)
\(=32+6-4076=-4038\)
x^3 + y^3 + z^3 = 3xyz
<=> (x + y + z)(x^2 + y^2 + z^2 -xy -yz - zx) = 0
vì x+y+z khác 0 => x^2 + y^2 + z^2 -xy -yz - zx = 0
nhân 2 vế cho 2 => (x - y)^2 + (y - z)^2 + (z -x)^2 = 0
=> x = y = z
thay vào P ta dc: P= xxx/(2x.2x.2x) = x^3/8x^3 = 1/8
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 : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)
dat a=x-y
b=y-z
c=z-x
a+b+c=0=x+y+z
\(\left(\frac{a}{z}+\frac{b}{x}+\frac{c}{y}\right)\left(\frac{z}{a}+\frac{x}{b}+\frac{y}{c}\right)\)
dung bumiakopsky de giai
...........................................
phân tích gt sau đó suy ra x+y+x=0
từ đây tính đc x+y=? y+z=? x+z=?
ta được kết quả là'; -2006
Xét \(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\left(x+y+z\right)\left(x^2+2xy+y^2-xy-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
TH1:\(x+y+z=0\)
\(\Rightarrow x+y=-z;y+z=-x;z+x=-y\left(1\right)\)
Thay (1) vô pt cần tính:
\(\frac{2016xyz}{-z.-x.-y}=\frac{2016xyz}{-\left(xyz\right)}=-2016\)
TH2:\(x^2+y^2+z^2-xy-yz-xz=0\)
Nhân 2 vế với 2
\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
Do VT dương
\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}}\Rightarrow x=y=z\)
Thay y,z ở pt cần tính là x
\(\Rightarrow\frac{2016x.x.x}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=\frac{2016}{8}=252\)
Vậy pt có thể = -2016 khi x + y + z = 0
pt có thể = 252 khi \(x^2+y^2+z^2-xy-xz-yz=0\)
\(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0\)
\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-xz-yz=0\end{cases}}\)
TH1 : \(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y-z=-x\end{cases}}\)
Thay vào biểu thức ta có :
\(\frac{2016xyz}{\left(-z\right)\left(-x\right)\left(-y\right)}=\frac{2016xyz}{-xyz}=-2016\)
TH2 : \(x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+\left(y^2-2yz+z^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}\Rightarrow}x=y=z}\)
Từ đây ta biến đổi biểu thức :
\(\frac{2016x^3}{2x\cdot2x\cdot2x}=\frac{2016x^3}{8x^3}=252\)
Vậy........
Ta có: \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x-y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\) hoặc \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)
-Nếu \(x+y+z=0\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)
Suy ra \(\frac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2016xyz}{-xyz}=-2016\)
-Nếu \(x=y=z\)
Suy ra: \(\frac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=252\)
Vậy ......