Esercizio
$\left(\frac{1}{2n}-\frac{3}{2}m\right)^4$
Soluzione passo-passo
Passi intermedi
1
Applicare la formula: $\left(a+b\right)^4$$=a^4+4a^3b+6a^2b^2+4ab^3+b^4$, dove $a=\frac{1}{2n}$, $b=-\frac{3}{2}m$ e $a+b=\frac{1}{2n}-\frac{3}{2}m$
$\left(\frac{1}{2n}\right)^4-6\left(\frac{1}{2n}\right)^3m+6\left(\frac{1}{2n}\right)^2\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$
Passi intermedi
2
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=2n$ e $n=4$
$\frac{1}{\left(2n\right)^4}-6\left(\frac{1}{2n}\right)^3m+6\left(\frac{1}{2n}\right)^2\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$
Passi intermedi
3
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=2n$ e $n=3$
$\frac{1}{\left(2n\right)^4}-6\left(\frac{1}{\left(2n\right)^3}\right)m+6\left(\frac{1}{2n}\right)^2\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$
Passi intermedi
4
Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=1$, $b=2n$ e $n=2$
$\frac{1}{\left(2n\right)^4}-6\left(\frac{1}{\left(2n\right)^3}\right)m+6\left(\frac{1}{\left(2n\right)^2}\right)\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$
Passi intermedi
5
Applicare la formula: $\left(ab\right)^n$$=a^nb^n$
$\frac{1}{16n^4}-6\left(\frac{1}{8n^3}\right)m+6\left(\frac{1}{4n^2}\right)\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$
Risposta finale al problema
$\frac{1}{16n^4}-6\left(\frac{1}{8n^3}\right)m+6\left(\frac{1}{4n^2}\right)\left(-\frac{3}{2}m\right)^2+\frac{2\left(-\frac{3}{2}m\right)^3}{n}+\left(-\frac{3}{2}m\right)^4$