Question

Найти значение (b-1/b)^2 если (b+1/b)^2=8

Answer 1

$\left(b+ \frac{1}{b}\right)^2=8$

$b^2+2 \cdot b \cdot \frac{1}{b} + \frac{1}{b^2}=8$

$b^2+2 + \frac{1}{b^2}=8$

$b^2+\frac{1}{b^2}=8-2$

$b^2+ \frac{1}{b^2}=6$

$\left(b- \frac{1}{b}\right)^2=b^2-2 \cdot b \cdot \frac{1}{b} + \frac{1}{b^2}=$

$b^2-2+ \frac{1}{b^2}=6-2=4$