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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)

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

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

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

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

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

thanks

Xét vế trái ta có: x^2 + y^2 + z^2 + 2xy + 2yz + 2xz

                       =x^2 + 2xy + y^2 + 2yz + 2xz +z^2

                       =(x+y)^2 + 2(x+y)z +z^2

                       =(x+y+z)^2

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

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

 =[(x+y)+z]²

=(x+y)²+2(x+y)z+z²

=x²+2xy+y²+2xz+2yz+z²

=x²+y²+z²+2xy+2yz+2xz

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

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