Question

Bài 1: CMR:

P= x.(x - y).(x + y). ( x + 2y) + y là bình phương của một đa thức.

Bài 2: Cho x + y + z = 0. CMR: (x² + y² + z²) = 2. x⁴ + y⁴ + z⁴

* Trả lời giúp mình vs. Mình đang cần gấp <3

Answer 1

x + y + z = 0 \(\Rightarrow\) x = - ( y + z )

\(\Rightarrow\) \(x^2\) = \((y+z)^2\) = \(y^2\) + \(z^2 \) + 2yz

\(\Rightarrow\) \(x^2\) - \(y^2\) - \(z^2 \) = 2xy

\(\Rightarrow\) (\(x^2-y^2-z^2\) )\(^2 \) = \((2xy)^2\)= \(4x^2y^2\)

\(\Rightarrow\) \(x^4 + y^4 + z^4\) - \(2x^2y^2\) - \(2x^2z^2\) = \(4x^2y^2\)

\(\Rightarrow\) \(x^4+y^4+z^4\) = \(4y^2z^2\) - \(2y^2z^2\) + \(2x^2y^2\) = \(2x^2y^2 + 2y^2z^2+ 2x^2z^2\)

\(\Rightarrow\) 2 (\(x^4+y^4+z^4\) ) = \((x^2+y^2+z^2)^2\) (đpcm)

Answer 2

\(x+y+z=0\Rightarrow x=-\left(y+z\right)\)

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

\(\Rightarrow x^2-y^2-z^2=2xy\)

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

\(\Rightarrow x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2=4x^2y^2\)

\(\Rightarrow x^4+y^4+x^4=4y^2z^2-2y^2z^2+2x^2z^2+2x^2y^2=2x^2y^2+2y^2z^2+2x^2z^2\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)