A sẽ giải thích tại sao đặt được nhân tử vậy cho nhé

Ta có:

\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

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

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

\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)

\(=\left(x-y\right)\left(\left(xy-zx\right)+\left(z^2-zy\right)\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Cậu ta làm sai thì làm sao ngonhuminh với thầy phynit hiểu được

1 đó

Từ \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) + \(\dfrac{1}{z}\) = 0

\(=>yz+xz+xy=0\)

\(=>yz=-xz-xy\)

Ta có : \(x^2+2yz=x^2+yz+yz=x^2+yz-yx-xz=\left(y-x\right)\left(z-x\right)=-\left(x-y\right)\left(z-x\right)\)

Tương tư :

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

\(z^2+2xy=z^2+xy+xy=z^2+xy-yz-xz=\left(z-y\right)\left(z-x\right)=-\left(y-z\right)\left(z-x\right)\)Nên A = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)

=\(\dfrac{-yz}{\left(x-y\right)\left(z-x\right)}+\dfrac{-xz}{\left(x-y\right)\left(y-z\right)}+\dfrac{-xy}{\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

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

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

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

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

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

=1

= =" ... bài này làm dài ..bấm máy mỏi tay lắm...

nhanh gọn lẹ.... A = 0

dài đấy

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ < =>xy+yz+xz=0\\ < =>\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-yz\end{matrix}\right.\)

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

cmtt

\(=>\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)

A = ...

= \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\)

=\(\dfrac{yz+xz+xy}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)

mà xy + yz + xz = 0

=> (1) = 0

=> a = 0

Pạn tham khảo cách làm nha!!!

Chúc pạn hok tốt!!!

Lời giải:

Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)

Suy ra \(yz=-xy-xz\)

\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)

\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)

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

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)

\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)

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

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

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)

\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)

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

Tương tự:

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)

Vậy \(A=0.\)