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

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

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

Áp dụng HĐT (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca đó bạn. 

Ta có: (x - y + z)^2 >= 0 
<=> x^2 + y^2 + z^2 - 2xy + 2xz - 2yz >= 0 
<=> x^2 + y^2 + z^2 >= 2xy - 2xz + 2yz

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