展开 ((n+1)(n)b-nc+c)/((n+2)(n+1))

解答

展开

\frac{( n + 1 ) ( n ) b - n c + c}{( n + 2 ) ( n + 1 )}

\frac{n ^{2} b + nb - n c + c}{n ^{2} + 3 n + 2}

求解步骤

\frac{( n + 1 ) ( n ) b - n c + c}{( n + 2 ) ( n + 1 )}

去除括号:

(a) = a

= \frac{( n + 1 ) nb - n c + c}{( n + 2 ) ( n + 1 )}

乘开

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

= \frac{nb ( n + 1 ) - n c + c}{n ^{2} + 3 n + 2}

乘开

(n + 1) nb - n c + c : n^{2} b + nb - n c + c

= \frac{n ^{2} b + nb - n c + c}{n ^{2} + 3 n + 2}

e x p an d \frac{1}{( s + 1 ) ( s + 0.7 )}

e x p an d - \frac{2}{3 ( x ^{2} + 4 ) ( x ^{2} + 9 )}

e x p an d \frac{2.4 + 0.7}{0.5} - \frac{3.6}{1.6}

\frac{d}{d x} ((f (x, y)) (3 x y - 4)^{2})

e x p an d \frac{1}{( x - 1 ) ( x - 3 ) ( 7 x ^{2} + 6 x + 1 )}