展開する x^2-(3x+(x(x+1)+4(x^2-1)-x^2)=0)

解

展開する

x^{2} - (3 x + (x (x + 1) + 4 (x^{2} - 1) - x^{2}) = 0)

x^{2} - 3 x - = 0

解答ステップ

x^{2} - (3 x + (x (x + 1) + 4 (x^{2} - 1) - x^{2}) = 0)

= x^{2} - (3 x + 0 = (x (x + 1) + 4 (x^{2} - 1) - x^{2}))

- (3 x + (x (x + 1) + 4 (x^{2} - 1) - x^{2}) = 0) : - 3 x - (x (x + 1) + 4 (x^{2} - 1) - x^{2}) = 0

x^{2} - 3 x - (x (x + 1) + 4 (x^{2} - 1) - x^{2}) = 0

拡張

- = 0 \cdot (x (x + 1) + 4 (x^{2} - 1) - x^{2}) : - = 0

x^{2} - 3 x - = 0