展開する-(2(x^2-5x+3))/(ln(5)(2x-1)(x+1)(x-1))

解

展開する

- \frac{2 ( x ^{2} - 5 x + 3 )}{ln ( 5 ) ( 2 x - 1 ) ( x + 1 ) ( x - 1 )}

- \frac{2 x ^{2} - 10 x + 6}{2 ln ( 5 ) x ^{3} - 2 ln ( 5 ) x - ln ( 5 ) x ^{2} + ln ( 5 )}

解答ステップ

- \frac{2 ( x ^{2} - 5 x + 3 )}{ln ( 5 ) ( 2 x - 1 ) ( x + 1 ) ( x - 1 )}

拡張

ln (5) (2 x - 1) (x + 1) (x - 1) : 2 ln (5) x^{3} - 2 ln (5) x - ln (5) x^{2} + ln (5)

= - \frac{2 ( x ^{2} - 5 x + 3 )}{2 ln ( 5 ) x ^{3} - ln ( 5 ) x ^{2} - 2 ln ( 5 ) x + ln ( 5 )}

拡張

2 (x^{2} - 5 x + 3) : 2 x^{2} - 10 x + 6

= - \frac{2 x ^{2} - 10 x + 6}{2 ln ( 5 ) x ^{3} - ln ( 5 ) x ^{2} - 2 ln ( 5 ) x + ln ( 5 )}