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17 tháng 7 2020

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

Tương tự:\(\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}=\frac{y}{x+y}-\frac{z}{y+z};\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=\frac{z}{z+y}-\frac{x}{z+x}\)

Khi đó:

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

5 tháng 7 2019

Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)

=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)

\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)

\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)

\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)

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

Như vậy:

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

<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)

Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)

 \(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)

\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)

13 tháng 11 2019

Giúp mình với các bạn

23 tháng 5 2015

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

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

                   lik.e nhé!

8 tháng 10 2018

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

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

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

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

\(=0\)\(\left(\text{Đ}K:x+y,y+z,z+x\ne0\right)\)

Tham khảo nhé~

5 tháng 11 2016

\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)

\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

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

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

Tới đây bạn thay vào nhé :)