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AH
Akai Haruma
Giáo viên
7 tháng 2 2020

Lời giải:

$x^3+y^3+z^3-3xyz=0$

$\Leftrightarrow (x+y)^3-3xy(x+y)+z^3-3xyz=0$

$\Leftrightarrow (x+y)^3+z^3-3xy(x+y+z)=0$

$\Leftrightarrow (x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)=0$

$\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0$

Đến đây xét 2TH:

TH1: $x+y+z=0$

\(\Rightarrow \left\{\begin{matrix} x+y=-z\\ y+z=-x\\ x+z=-y\end{matrix}\right.\)

\(\Rightarrow B=-16+(-3)+(-2038)=-2057\)

TH2: $x^2+y^2+z^2-xy-yz-xz=0$

$\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0$

$\Rightarrow (x-y)^2=(y-z)^2=(z-x)^2=0$

$\Rightarrow x=y=z$ (vô lý vì $x,y,z$ đôi một khác nhau)

Vậy.......

NV
7 tháng 2 2020

\(x^3+y^3+z^3-3xyz=0\Leftrightarrow x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)

- Nếu \(x+y+z=0\Rightarrow B=\frac{-16z}{z}-\frac{3x}{x}-\frac{2038y}{y}=...\)

- Nếu \(x=y=z\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}+\frac{2038.2y}{y}=...\)

(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

7 tháng 2 2020

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 \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)

\(=32+6-4076=-4038\)

NV
12 tháng 3 2021

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)

6 tháng 7

GIÁO VIÊN SAO TOÀN SAI HẰNG ĐẲNG THỨC THẾ????

25 tháng 7 2019

\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)

\(A=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{y+z}{z}\)

Do \(x-y-z=0\)

\(\Rightarrow x-z=y;y-x=-z;y+z=x\)

Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)

Vậy A=-1

25 tháng 7 2019

\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)

\(=\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)

\(=\frac{1}{xy+x+1}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz+y+1}{yz+y+1}\)

\(=1\)

11 tháng 4 2017

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

...........................................

30 tháng 8 2021

áp dụng tc của dãy tỉ số = nhau : 

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)

\(\Rightarrow\hept{\begin{cases}y+z-x=x\\z+x-y=y\\x+y-z=z\end{cases}\Leftrightarrow\hept{\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}z-x=2x-2z\\y-x=2x-2y\\z-y=2y-z\end{cases}\Leftrightarrow\hept{\begin{cases}3x=3z\\3x=3y\\3y=3z\end{cases}}\Leftrightarrow x=y=z}\)

thay vào B ta đc : \(B=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=8\)

30 tháng 8 2021

Ta có : \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)

=> \(\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)

=> \(\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)

Khi x + y + z = 0 

=> x + y = -z ; y + z = -x ; z + x = -y

Khi đó \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}=\frac{-z.\left(-x\right).\left(-y\right)}{y.z.x}=-1\)

Khi x  + y + z \(\ne\)0

=> x = y = z 

Khi đó \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

11 tháng 1 2021

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 : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)

11 tháng 1 2021

Khi x + y + z = 0

=> x + y = -z

=> x + z = - y

=> y + z = - x

Khi đó M = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)

12 tháng 10 2016

Đặt \(\sqrt{x}=a\) , \(\sqrt{y}=b\) , \(\sqrt{z}=c\)

Suy ra \(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Xét tử : \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left[-\left(a-b\right)-\left(c-a\right)\right]+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c^2-a^2\right)+\left(c-a\right)\left(b^2-a^2\right)=\left(a-b\right)\left(c-a\right)\left(c+a\right)+\left(c-a\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c+a-a-b\right)=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

Suy ra \(P=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)