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

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

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

(x²+y²)²-(2xy)²

=(x²+y²+2xy)(x²+y²-2xy)

=(x²+2xy+y²)(x²-2xy+y²)

=(x+y)²(x-y)²

Ta có: \(\left(x-y-z\right)^2\)

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

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

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

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

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

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

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

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

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

Ta có : y² = xy \(\Rightarrow\)x = y  ( 1 )

x² = yz hay x² = xz \(\Rightarrow\)x = z ( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)x = y = z

Vậy x = y = z

a) (x + y)² = (x + y)(x + y) = x² + xy + xy + y² = x² + 2xy + y² (đpcm)

b) (x - y)² = (x - y)(x - y) = x² - xy - xy + y² = x² - 2xy + y² (đpcm)

a) Ta có: \(VT=\left(x+y\right)^2\)

\(=\left(x+y\right)\cdot\left(x+y\right)\)

\(=x^2+xy+yx+y^2\)

\(=x^2+2xy+y^2=VP\)(đpcm)

b) Ta có: \(VP=x^2-2xy+y^2\)

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

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

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

\(=\left(x-y\right)^2=VT\)(đpcm)

a)  \(\left(x-y\right)^2+2xy\)

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

\(=x^2+y^2\left(đpcm\right)\)

b)  \(\left(x-y\right)^2+4xy\)

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

\(=x^2+2xy+y^2\)

\(=\left(x+y\right)^2\left(đpcm\right)\)

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

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

b,Ta có:\(\left(x+y\right)^2=x^2+2xy+y^2\)

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

\(\Leftrightarrow\left(x+y\right)^2=\left(x-y\right)^2+4xy\left(ĐCCM\right)\)

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)

Có: \(\left(x+y\right)^2\ge0\)

\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)

\(\frac{1}{2}>0\)

\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x

xét vế trái:     \(x^2+2xy+2y^2+y+\frac{1}{2}\)    =\(x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

                                                                        = \(\left(x^2+2xy+y^2\right)+\left(y^2+y+\frac{1}{2}\right)\)

                                                                        = \(\left(x+y\right)^2+\left(y^2+2.\frac{1}{2}.y+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}\right)\)

                                                                        =  \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}\)

  vì \(\left(x+y\right)^2>=0\) và \(\left(y+\frac{1}{2}\right)^2>=0\)  =>   \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2>=0\)

   mà    1/4 >0    =>     \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}>0\)