Завдання № 9.27

№ 9.27 Алгебра = № 19.27 Математика

Спростіть вираз

$\left(1-x^{-2}\right)\cdot$

$\cdot \left(1-\frac{1}{x^{-1}-1}+\frac{1}{x^{-1}+1}\right);$

$\left(1-x^{-2}\right)\cdot$

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

$= \frac{3x^2-1}{x^2};$

1. $\ 1-x^{-2}=1-\frac{1}{x^2}=\frac{x^2-1}{x^2};$

2. $\ x^{-1}-1=\frac{1}{x}-1=\frac{1-x}{x};$

3. $\frac{1}{\frac{1-x}{x}}=\frac{x}{1-x};$

4. $\ x^{-1}+1=\frac{1}{x}+1=\frac{1+x}{x};$

5. $\frac{1}{\frac{1+x}{x}}=\frac{x}{1+x};$

6. $\ 1-\frac{x}{1-x}+\frac{x}{1+x}=$

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

$= \frac{1-x^2-x-x^2+x-x^2}{\left(1-x\right)\left(1+x\right)}=\frac{-3x^2+1}{1-x^2};$

7. $\frac{x^2-1}{x^2}\cdot\frac{-3x^2+1}{1-x^2}=$

$= -\frac{\left(x^2-1\right)\left(3x^2-1\right)}{x^2\left(1-x^2\right)}=$

$= \frac{\left(1-x^2\right)\left(3x^2-1\right)}{x^2\left(1-x^2\right)}= \frac{3x^2-1}{x^2}.$