it

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popolare Algebra >

espandere (1/4+3/4 e^t)^5

Soluzione

espandere

(\frac{1}{4} + \frac{3}{4} e^{t})^{5}

Soluzione

\frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{243 e ^{5 t}}{1024} + \frac{45 e ^{2 t}}{512} + \frac{135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t}

Fasi della soluzione

(\frac{1}{4} + \frac{3}{4} e^{t})^{5}

Applicare il teorema binomiale:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = \frac{1}{4}, b = \frac{3}{4} e^{t}

= i = 0 \sum 5 (i 5) (\frac{1}{4})^{(5 - i)} (\frac{3}{4} e^{t})^{i}

Espandere la sommatoria

= \frac{5 !}{0 ! ( 5 - 0 ) !} (\frac{1}{4})^{5} (\frac{3}{4} e^{t})^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} (\frac{1}{4})^{4} (\frac{3}{4} e^{t})^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} (\frac{1}{4})^{3} (\frac{3}{4} e^{t})^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} (\frac{1}{4})^{2} (\frac{3}{4} e^{t})^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} (\frac{1}{4})^{1} (\frac{3}{4} e^{t})^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} (\frac{1}{4})^{0} (\frac{3}{4} e^{t})^{5}

\frac{5 !}{0 ! ( 5 - 0 ) !} (\frac{1}{4})^{5} (\frac{3}{4} e^{t})^{0} = \frac{1}{1024}

Semplifica

\frac{5 !}{1 ! ( 5 - 1 ) !} (\frac{1}{4})^{4} (\frac{3}{4} e^{t})^{1} : \frac{15 e ^{t}}{1024}

Semplifica

\frac{5 !}{2 ! ( 5 - 2 ) !} (\frac{1}{4})^{3} (\frac{3}{4} e^{t})^{2} : \frac{45 e ^{2 t}}{512}

Semplifica

\frac{5 !}{3 ! ( 5 - 3 ) !} (\frac{1}{4})^{2} (\frac{3}{4} e^{t})^{3} : \frac{135 e ^{3 t}}{512}

Semplifica

\frac{5 !}{4 ! ( 5 - 4 ) !} (\frac{1}{4})^{1} (\frac{3}{4} e^{t})^{4} : \frac{405}{1024} e^{4 t}

Semplifica

\frac{5 !}{5 ! ( 5 - 5 ) !} (\frac{1}{4})^{0} (\frac{3}{4} e^{t})^{5} : \frac{243 e ^{5 t}}{1024}

= \frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{45 e ^{2 t}}{512} + \frac{135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t} + \frac{243 e ^{5 t}}{1024}

Semplifica

\frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{45 e ^{2 t}}{512} + \frac{135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t} + \frac{243 e ^{5 t}}{1024} : \frac{1 + 15 e ^{t} + 243 e ^{5 t}}{1024} + \frac{45 e ^{2 t} + 135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t}

= \frac{1 + 15 e ^{t} + 243 e ^{5 t}}{1024} + \frac{45 e ^{2 t} + 135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t}

Applica la regola delle frazioni:

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

\frac{1 + 15 e ^{t} + 243 e ^{5 t}}{1024} = \frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{243 e ^{5 t}}{1024}

= \frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{243 e ^{5 t}}{1024} + \frac{45 e ^{2 t} + 135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t}

Applica la regola delle frazioni:

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

\frac{45 e ^{2 t} + 135 e ^{3 t}}{512} = \frac{45 e ^{2 t}}{512} + \frac{135 e ^{3 t}}{512}

= \frac{1}{1024} + \frac{15 e ^{t}}{1024} + \frac{243 e ^{5 t}}{1024} + \frac{45 e ^{2 t}}{512} + \frac{135 e ^{3 t}}{512} + \frac{405}{1024} e^{4 t}

Esempi popolari

espandere 1/(s^3(s^2-1))

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

espandere (x+y)(x^'+y+z)

e x p an d (x + y) (x^{^{'}} + y + z)

espandere (((z_{1}+z_{2})^2))/(z_{3)}

e x p an d \frac{( ( z _{1} + z _{2} ) ^{2} )}{z _{3}}

espandere-1/12 (x)(((96-16x))/(12))^3

e x p an d - \frac{1}{12} (x) (\frac{( 96 - 16 x )}{12})^{3}

espandere (x(n+1)(n^2+3))/(n(n^2+2n+4))

e x p an d \frac{x ( n + 1 ) ( n ^{2} + 3 )}{n ( n ^{2} + 2 n + 4 )}