The fraction $\frac{a^{2} + b^{2} - c^{2} + 2 a b}{a^{2} + c^{2} - b^{2} + 2 a c}$ is (with suitable restrictions on the values of $a, b,$ and $c$ ).

Irreducible

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Reducible to

- 1

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Reducible to a polynomial of three terms

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Reducible to

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

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Reducible to

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

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Solution

The correct option is E Reducible to $\frac{a + b - c}{a - b + c}$
$\frac{a^{2} + b^{2} - c^{2} + 2 a b}{a^{2} + c^{2} - b^{2} + 2 a c} = \frac{(a + b)^{2} - c^{2}}{(a + c)^{2} - b^{2}} = \frac{(a + b + c) (a + b - c)}{(a + c + b) (a + c - b)}$
$= \frac{a + b - c}{a + c - b}$ with $(a + c)^{2} \neq b^{2}$ .

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Similar questions

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}}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

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

Divide:
$[(a^{2} + 2 a b + b^{2}) - (a^{2} + 2 a c + c^{2})]$ by $2 a + b + c$

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