What is (1/x+y)(1/x-y) ?

The solution to (1/x+y)(1/x-y) is ((yx+1)(-yx+1))/(x^2)

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simplify (1/x+y)(1/x-y)

simplify

(\frac{1}{x} + y) (\frac{1}{x} - y)

\frac{( y x + 1 ) ( - y x + 1 )}{x ^{2}}

Solution steps

(\frac{1}{x} + y) (\frac{1}{x} - y)

Join

\frac{1}{x} + y : \frac{y x + 1}{x}

= \frac{y x + 1}{x} (\frac{1}{x} - y)

Apply the fraction rule:

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

= \frac{( y x + 1 ) ( \frac{1}{x} - y )}{x}

Simplify

(y x + 1) (\frac{1}{x} - y) : \frac{( y x + 1 ) ( - y x + 1 )}{x}

= \frac{\frac{( y x + 1 ) ( - y x + 1 )}{x}}{x}

Apply the fraction rule:

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

= \frac{( y x + 1 ) ( - y x + 1 )}{xx}

Apply exponent rule:

aa = a^{2}

xx = x^{2}

= \frac{( y x + 1 ) ( - y x + 1 )}{x ^{2}}

(e^{x i} - cos (x))^{2}

(\frac{6}{7} - x)^{5}

\frac{\partial}{\partial x} ((x) (x - 3) (y - 4))

f (x) x (2 (y)^{''} - x (y)^{'} = e^{- x})

e^{x} (x + e^{x})

What is (1/x+y)(1/x-y) ?
The solution to (1/x+y)(1/x-y) is ((yx+1)(-yx+1))/(x^2)