a ) Đặt \(x^2=a;y^2=b\) . Khi đó , ta có :
\(\dfrac{1}{2}\left(a+b\right)^2-2ab\)
\(=\dfrac{\left(a+b\right)^2-4ab}{2}\)
\(=\dfrac{a^2+b^2+2ab-4ab}{2}\)
\(=\dfrac{a^2+b^2-2ab}{2}\)
\(=\dfrac{\left(a-b\right)^2}{2}\)
\(=\dfrac{\left(x^2-y^2\right)^2}{2}\)
\(=\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\)
b ) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left[x-y+y-z\right]+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(x-y\right)-y^2\left(y-z\right)+z^2\left(x-y\right)\)
\(=\left(x^2-y^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[x+y-\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
