Ta có
\(\left(x^2+y^2+z^2\right)^2-2\left(x^4+y^4+z^4\right)\)
\(=2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4\)
\(=\left(z^2x^2+2z^2xy+z^2y^2\right)+\left(z^2x^2-2z^2xy+z^2y^2\right)+\left(-x^4+2x^2y^2-y^4\right)-z^4\)
\(=z^2\left(x+y\right)^2+z^2\left(x-y\right)^2-\left(x^2-y^2\right)^2-z^4\)
\(=z^2\left(\left(x+y\right)^2-z^2\right)-\left(x-y\right)^2\left(\left(x+y\right)^2-z^2\right)\)
\(=\left(\left(x+y\right)^2-z^2\right)\left(z^2-\left(x-y\right)^2\right)\)
\(=\left(x+y+z\right)\left(x+y-z\right)\left(z-x+y\right)\left(z+x-y\right)=0\)
Vậy \(\left(x^2+y^2+z^2\right)^2=2\left(x^4+y^4+z^4\right)\)
x+y+z=0
=> x=-(y+z) => x2=y2+2yz+z2
=> 2yz=x2-y2-z2=> 4y2z2=x4+y4+z4-2x2y2-2x2z2+2y2z2
=> 2x2y2+2x2z2+2y2z2= x4+y4+z4 (1)
mặt khác (x2+y2+z2)2=x4=y4+z4+2x2y2+2x2z2+2y2z2 (2)
từ (1)(2) ta được (x2+y2+z2)2=2(x4+y4+z4)
2x2y2+2y2z2+2z2x2−x4−y4−z4
=2x2y2+2y2z2+2z2x2−x4−y4−z
=(z2x2+2z2xy+z2y2)+(z2x2−2z2xy+z2y2)+(−x4+2x2y2−y4)−z4
=z2(x+y)2+z2(x−y)2−(x2−y2)2−z4
=((x+y)2−z2)(z2−(x−y)2)
=(x+y+z)(x+y−z)(z−x+y)(z+x−y)=0
Vậy (x2+y2+z2)2=2(x4+y4+z4)
Đúng 7
Bạn có thể tham khảo qua cách mình:
Ta có: \(x+y+z=0\Rightarrow x=-\left(y+z\right)\Rightarrow x^2=\left(y+z\right)^2\)
\(\left(x^2+y^2+z^2\right)^2-2\left(x^4+y^4+z^4\right)\)
=\(x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)-2\left(x^4+y^4+z^4\right)\)
=\(-x^4-y^4-z^4+2x^2y^2+2y^2z^2+2z^2x^2\)
=\(-x^2\left(y+z\right)^2-y^2\left(x+z\right)^2-z^2\left(x+y\right)^2+2x^2y^2+2y^2z^2+2z^2x^2\)
=\(\left(-x^2y^2-2x^2yz-x^2z^2\right)+\left(-y^2z^2-2y^2xz-y^2x^2\right)+\left(-z^2x^2-2z^2xy-z^2y^2\right)+2x^2y^2+2y^2z^2+2z^2x^2\)
=\(-2x^2yz-2y^2xz-2z^2xy\)
=\(-2xyz\left(x+y+z\right)\)
=\(0\)
Vậy ta có đpcm
ttttruogjzjzbsjxgvgxd/v
cậu học lớp mấy mà chữ tớ chẳng hiểu gì cả
