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10 tháng 2 2019

đặt \(\frac{x-y}{z}=a;\frac{y-z}{x}=b;\frac{z-x}{y}=c\)

\(\Rightarrow\)\(\frac{z}{x-y}=\frac{1}{a};\frac{x}{y-z}=\frac{1}{b};\frac{y}{z-x}=\frac{1}{c}\)

Ta có : \(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(A=1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+1=3+\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

Ta có :  \(\frac{b+c}{a}=\left(b+c\right)\frac{1}{a}=\left(\frac{y-z}{x}+\frac{z-x}{y}\right)\frac{z}{x-y}=\frac{y^2-yz+xz-x^2}{xy}.\frac{z}{x-y}=\frac{\left(y-x\right)\left(x+y-z\right)}{xy}.\frac{z}{x-y}=\frac{\left(z-x-y\right)z}{xy}=\frac{2z^2}{xy}\)vì x + y + z = 0 \(\Rightarrow\)z = -x - y

Tương tự : \(\frac{a+c}{b}=\frac{2x^2}{yz}\)\(\frac{a+b}{c}=\frac{2y^2}{xz}\)

\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2z^2}{xy}+\frac{2x^2}{yz}+\frac{2y^2}{xz}=\frac{2\left(x^3+y^3+z^3\right)}{xyz}=\frac{2.3xyz}{xyz}=6\)( vì x + y + z = 0 \(\Rightarrow\)x3 + y3 + z3 = 3xyz )

Vậy A = 3 + 6 = 9

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

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12 tháng 2 2020

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0

\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)

Tương tự rồi cộng lại thì \(BT=0\) thì phải

12 tháng 2 2020

Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)

Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)

\(=\frac{1}{x^2+\left(y-z\right)\left(y+z\right)}+\frac{1}{y^2+\left(z-x\right)\left(z+x\right)}+\frac{1}{z^2+\left(x-y\right)\left(x+y\right)}\left(4\right)\)

Because \(x^2+y^2+z^2=3xyz\)

\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)

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

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

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)

The first case: If \(x+y+z=0\left(1\right)\)

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

From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)

 \(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have

\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)

\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)

\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)

\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)

The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)

We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)

From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)

Because \(x=y=z\Rightarrow x^2=y^2=z^2=xy=yz=zx\)

So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

\(=\frac{z+x+y}{xyz}=0\)

So...

28 tháng 12 2015

=>(x+y)(z-x)=(x+z)(x-y)

x(z-x)+y(z-x)=x(x-y)+z(x-y)

zx-x^2+yz-xy=x^2-xy+zx-yz

(yz+yz)+(zx-zx)=(x^2+x^2)-(xy-xy)

2yz=2x^2

=>yz=x^2

nên x^2-yz=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

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

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

27 tháng 9 2015

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

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

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

Suy ra: \(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}\)

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

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

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