(x + y + z)2 – 2.(x + y + z).(x + y) + (x + y)2

= [(x + y + z) – (x + y)]2 (Áp dụng HĐT (2) với A = x + y + z ; B = x + y)

= z2.

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

Sai đề.

Ta có: x+y+z=0

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

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)(1)

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

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

\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}\)

\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2yz-2xz\right)}\)

\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

Vậy: \(K=\dfrac{1}{3}\)

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

\(K=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\dfrac{1}{3}\)

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

b) \(=2\left(x^2-y^2\right)+2\left(x^2+y^2\right)=2x^2+2x^2+2y^2-2y^2=4x^2\)( cái này áp dụng luôn kết quả câu trên nha)

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

tớ cũng giống Nguyễn Thị Bích Hậu

tích cho nha 1 cái thôi cũng được .

\(\left(a\right):\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\)

\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)

\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)