expandir (4r-5a)^4

Solução

expandir

(4 r - 5 a)^{4}

256 r^{4} - 1280 r^{3} a + 2400 r^{2} a^{2} - 2000 r a^{3} + 625 a^{4}

Passos da solução

(4 r - 5 a)^{4}

Aplicar o teorema do binômio:

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

a = 4 r, b = - 5 a

= i = 0 \sum 4 (i 4) (4 r)^{(4 - i)} (- 5 a)^{i}

Expandir a soma

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

Simplificar

\frac{4 !}{0 ! ( 4 - 0 ) !} (4 r)^{4} (- 5 a)^{0} : 256 r^{4}

Simplificar

\frac{4 !}{1 ! ( 4 - 1 ) !} (4 r)^{3} (- 5 a)^{1} : - 1280 r^{3} a

Simplificar

\frac{4 !}{2 ! ( 4 - 2 ) !} (4 r)^{2} (- 5 a)^{2} : 2400 r^{2} a^{2}

Simplificar

\frac{4 !}{3 ! ( 4 - 3 ) !} (4 r)^{1} (- 5 a)^{3} : - 2000 r a^{3}

Simplificar

\frac{4 !}{4 ! ( 4 - 4 ) !} (4 r)^{0} (- 5 a)^{4} : 625 a^{4}

= 256 r^{4} - 1280 r^{3} a + 2400 r^{2} a^{2} - 2000 r a^{3} + 625 a^{4}

e x p an d (a + b)^{3} - a - b

s im pl i f y (1 + \frac{x}{2} - \frac{2}{x})^{4}

s im pl i f y \frac{4}{( x )}

f a c t or 12 c - 28 d + 52

s im pl i f y (\frac{0.5 - 2 x}{0.5})^{2}