$\frac{a^{2} - b^{2} - 2 b c - c^{2}}{a^{2} + b^{2} + 2 a b - c^{2}}$ is equivalent to

\frac{a + b + c}{a - b + c}

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\frac{a - b - c}{a + b - c}

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\frac{a - b - c}{a - b + c}

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\frac{a - b + c}{a + b + c}

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Solution

The correct option is B $\frac{a - b - c}{a + b - c}$
$\frac{a^{2} - b^{2} - 2 b c - c^{2}}{a^{2} + b^{2} + 2 a b - c^{2}} = \frac{a^{2} - (b^{2} + 2 b c + c^{2})}{(a^{2} + b^{2} + 2 a b) - c^{2}} = \frac{a^{2} - (b + c)^{2}}{(a + b)^{2} - c^{2}} (∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b) = \frac{(a + b + c) (a - (b + c))}{(a + b + c) (a + b - c)} (∵ a^{2} - b^{2} = (a + b) (a - b))$
$= \frac{(a + b + c) (a - b - c)}{(a + b + c) (a + b - c)} = \frac{a - b - c}{a + b - c}$

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Similar questions

Find:
$(a + b + c) \times (a + b - c) =$

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

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} \geq a + b + c

a, b, c

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

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

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