展开 (125(2n+3)^2(2n+4)^2)/4

解答

展开

\frac{125 ( 2 n + 3 ) ^{2} ( 2 n + 4 ) ^{2}}{4}

500 n^{4} + 3500 n^{3} + 9125 n^{2} + 10500 n + 4500

求解步骤

\frac{125 ( 2 n + 3 ) ^{2} ( 2 n + 4 ) ^{2}}{4}

(2 n + 3)^{2} = 4 n^{2} + 12 n + 9

= \frac{125 ( 4 n ^{2} + 12 n + 9 ) ( 2 n + 4 ) ^{2}}{4}

(2 n + 4)^{2} = 4 n^{2} + 16 n + 16

= \frac{125 ( 4 n ^{2} + 12 n + 9 ) ( 4 n ^{2} + 16 n + 16 )}{4}

分解

125 (4 n^{2} + 12 n + 9) (4 n^{2} + 16 n + 16) : 500 (4 n^{2} + 12 n + 9) (n^{2} + 4 n + 4)

= \frac{500 ( 4 n ^{2} + 12 n + 9 ) ( n ^{2} + 4 n + 4 )}{4}

数字相除:

\frac{500}{4} = 125

= 125 (4 n^{2} + 12 n + 9) (n^{2} + 4 n + 4)

乘开

(4 n^{2} + 12 n + 9) (n^{2} + 4 n + 4) : 4 n^{4} + 28 n^{3} + 73 n^{2} + 84 n + 36

= 125 (4 n^{4} + 28 n^{3} + 73 n^{2} + 84 n + 36)

乘开

125 (4 n^{4} + 28 n^{3} + 73 n^{2} + 84 n + 36) : 500 n^{4} + 3500 n^{3} + 9125 n^{2} + 10500 n + 4500

= 500 n^{4} + 3500 n^{3} + 9125 n^{2} + 10500 n + 4500

e x p an d (2 π - 45)^{2}

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e x p an d (k + 1)! (k + 2)