簡約 (x+1)(x+2)+(x+3)(x+4)-2x^2-10x+1

解

簡約

(x + 1) (x + 2) + (x + 3) (x + 4) - 2 x^{2} - 10 x + 1

15

解答ステップ

(x + 1) (x + 2) + (x + 3) (x + 4) - 2 x^{2} - 10 x + 1

拡張

(x + 1) (x + 2) : x^{2} + 3 x + 2

= x^{2} + 3 x + 2 + (x + 3) (x + 4) - 2 x^{2} - 10 x + 1

拡張

(x + 3) (x + 4) : x^{2} + 7 x + 12

= x^{2} + 3 x + 2 + x^{2} + 7 x + 12 - 2 x^{2} - 10 x + 1

条件のようなグループ

= x^{2} + x^{2} - 2 x^{2} + 3 x + 7 x - 10 x + 2 + 12 + 1

類似した元を足す:

x^{2} + x^{2} - 2 x^{2} = 0

= 3 x + 7 x - 10 x + 2 + 12 + 1

類似した元を足す:

3 x + 7 x - 10 x = 0

= 2 + 12 + 1

数を足す:

2 + 12 + 1 = 15

= 15