espandere ((x^3+x)^{10})/(10)

Soluzione

espandere

\frac{( x ^{3} + x ) ^{10}}{10}

\frac{x ^{30}}{10} + x^{28} + \frac{9 x ^{26}}{2} + 12 x^{24} + 21 x^{22} + \frac{126 x ^{20}}{5} + 21 x^{18} + 12 x^{16} + \frac{9 x ^{14}}{2} + x^{12} + \frac{x ^{10}}{10}

Fasi della soluzione

\frac{( x ^{3} + x ) ^{10}}{10}

(x^{3} + x)^{10} = x^{30} + 10 x^{28} + 45 x^{26} + 120 x^{24} + 210 x^{22} + 252 x^{20} + 210 x^{18} + 120 x^{16} + 45 x^{14} + 10 x^{12} + x^{10}

= \frac{x ^{30} + 10 x ^{28} + 45 x ^{26} + 120 x ^{24} + 210 x ^{22} + 252 x ^{20} + 210 x ^{18} + 120 x ^{16} + 45 x ^{14} + 10 x ^{12} + x ^{10}}{10}

Applica la regola delle frazioni:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{x ^{30} + 10 x ^{28} + 45 x ^{26} + 120 x ^{24} + 210 x ^{22} + 252 x ^{20} + 210 x ^{18} + 120 x ^{16} + 45 x ^{14} + 10 x ^{12} + x ^{10}}{10} = \frac{x ^{30}}{10} + \frac{10 x ^{28}}{10} + \frac{45 x ^{26}}{10} + \frac{120 x ^{24}}{10} + \frac{210 x ^{22}}{10} + \frac{252 x ^{20}}{10} + \frac{210 x ^{18}}{10} + \frac{120 x ^{16}}{10} + \frac{45 x ^{14}}{10} + \frac{10 x ^{12}}{10} + \frac{x ^{10}}{10}

= \frac{x ^{30}}{10} + \frac{10 x ^{28}}{10} + \frac{45 x ^{26}}{10} + \frac{120 x ^{24}}{10} + \frac{210 x ^{22}}{10} + \frac{252 x ^{20}}{10} + \frac{210 x ^{18}}{10} + \frac{120 x ^{16}}{10} + \frac{45 x ^{14}}{10} + \frac{10 x ^{12}}{10} + \frac{x ^{10}}{10}

\frac{10 x ^{28}}{10} = x^{28}

\frac{45 x ^{26}}{10} = \frac{9 x ^{26}}{2}

\frac{120 x ^{24}}{10} = 12 x^{24}

\frac{210 x ^{22}}{10} = 21 x^{22}

\frac{252 x ^{20}}{10} = \frac{126 x ^{20}}{5}

\frac{210 x ^{18}}{10} = 21 x^{18}

\frac{120 x ^{16}}{10} = 12 x^{16}

\frac{45 x ^{14}}{10} = \frac{9 x ^{14}}{2}

\frac{10 x ^{12}}{10} = x^{12}

= \frac{x ^{30}}{10} + x^{28} + \frac{9 x ^{26}}{2} + 12 x^{24} + 21 x^{22} + \frac{126 x ^{20}}{5} + 21 x^{18} + 12 x^{16} + \frac{9 x ^{14}}{2} + x^{12} + \frac{x ^{10}}{10}

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

e x p an d \frac{( 4 x - 1 ) ( x + 5 )}{x - 4}

e x p an d 3 sin (4 (x - 5)) + 3

e x p an d 4 (x e^{2 x} + e^{x})

e x p an d (e^{2 x - 1}) (e^{x} + e^{- x})