x² +y² +z² -2xy-2zx-2yz=(x-y-z)² -4yz=(x-y-z)² - \(2.\sqrt{yz^2}\)=\(\left(x-y-z-2\sqrt{yz}\right)+\left(x-y-z+2\sqrt{yz}\right)\)

x² -2xy - y² -z² =(x-y)² -z² =(x-y-z)(x-y+z)

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

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

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

\(=\dfrac{x-y+z}{x-y-z}\)

.

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

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

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

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

Lời giải:

Từ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

$\Rightarrow xy+yz+xz=0$

Khi đó:

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

Tương tự với $y^2+2zx, z^2+2xy$ thì:

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

$=\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\frac{-[yz(y-z)+xz(z-x)+xy(x-y)]}{-[xy(x-y)+yz(y-z)+xz(z-x)]}=1$

a) Ta có: \(VT=\left(x-y-z\right)^2\)

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

\(=x^2-xy-xz-yx+y^2+yz-zx+zy+z^2\)

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

=VP(đpcm)

b) Ta có: \(VT=\left(x+y-z\right)^2\)

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

\(=x^2+xy-xz+yx+y^2-yz-zx-zy+z^2\)

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

=VP(đpcm)

c) Sửa đề: Chứng minh \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\)

Ta có: \(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=x^4-y^4\)

=VP(đpcm)

d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)

\(=x^5+y^5\)

=VP(đpcm)

a, b, nhân vào là ra à

c, nghe cứ là lạ

d, cũng nhân là ra hà

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5\)