✨ Best Answer ✨
(3)左の( )から①②③④ね
(x+y+z)(x+y-z)=(x+y)²-z²←①と④
(-x+y+z)(x-y+z)={z-(x-y)}{z+(x-y)}=z²-(x-y)²←②と③
よって、
(x²+y²+2xy-z²)(z²-x²+2xy-y²)
={2xy+(x²+y²-z²)}{2xy-(x²+y²-z²)}
=(2xy)²-(x²+y²-z²)²
後の計算は頑張って。もっと楽な方法あるかもしれませんが
(3)と(4)が分からん(泣)
助けてくださいヽ( ̄д ̄;)ノ=3=3=3
✨ Best Answer ✨
(3)左の( )から①②③④ね
(x+y+z)(x+y-z)=(x+y)²-z²←①と④
(-x+y+z)(x-y+z)={z-(x-y)}{z+(x-y)}=z²-(x-y)²←②と③
よって、
(x²+y²+2xy-z²)(z²-x²+2xy-y²)
={2xy+(x²+y²-z²)}{2xy-(x²+y²-z²)}
=(2xy)²-(x²+y²-z²)²
後の計算は頑張って。もっと楽な方法あるかもしれませんが
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ありがとうございます!計算頑張るぅ(*´∀`*)