What is ((b^2-c^2)/(bc)}{\frac{b-c)/c} ?

The solution to ((b^2-c^2)/(bc)}{\frac{b-c)/c} is (b+c)/b

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simplify ((b^2-c^2)/(bc)}{\frac{b-c)/c}

simplify

\frac{\frac{b ^{2} - c ^{2}}{b c}}{\frac{b - c}{c}}

\frac{b + c}{b}

Solution steps

\frac{\frac{b ^{2} - c ^{2}}{b c}}{\frac{b - c}{c}}

Apply the fraction rule:

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

= \frac{( b ^{2} - c ^{2} ) c}{b c ( b - c )}

Cancel the common factor:

c

= \frac{b ^{2} - c ^{2}}{b ( b - c )}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

b^{2} - c^{2} = (b + c) (b - c)

= \frac{( b + c ) ( b - c )}{b ( b - c )}

Cancel the common factor:

b - c

= \frac{b + c}{b}

- 7 \times 7 + 9 \times \frac{- 1}{- 1}

\frac{n - 2}{5} - \frac{n}{3} \leq 2

m^{2} = - 5

x^{2} + \frac{3}{8} x + 2 = 0

\frac{4 ( - 1 )}{3 + 2 ( 1 ) ^{2} + 2 ( - 1 ) ^{2}}

What is ((b^2-c^2)/(bc)}{\frac{b-c)/c} ?
The solution to ((b^2-c^2)/(bc)}{\frac{b-c)/c} is (b+c)/b