Помогите с решением.......................

$\frac{ \frac{1}{a} - \frac{1}{b + c} }{ \frac{1}{a} + \frac{1}{b + c} } \times (1 + \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} ) \div \frac{a - b - c}{abc} = \\ = \frac{ \frac{b + c - a}{a(b + c)} }{ \frac{b + c + a}{a(b + c)} } \times \frac{2bc + {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \times \frac{abc}{a - b - c} = \\ = \frac{b + c - a}{a(b + c)} \times \frac{a(b + c)}{b + c + a} \times \frac{( {b}^{2} + 2bc + {c}^{2}) - {a}^{2} }{2} \times \frac{a}{a - b - c} = \\ = \frac{b + c - a}{b + c + a} \times \frac{(b + c)^{2} - {a}^{2} }{2} \times \frac{a}{a - b - c} = \\ = \frac{b + c - a}{b + c + a} \times \frac{(b + c - a )(b + c + a)}{2} \times \frac{a}{a - b - c} = \\ = \frac{b + c - a}{1} \times \frac{b + c - a }{2} \times \frac{a}{ - ( - a + b + c)} = \\ = \frac{ {(b + c - a)}^{2} }{ 2} \times \frac{a}{ - (b + c - a)} = \\ = \frac{a(b + c - a)}{ - 2}$
a=0,02
b=-11,05
c=1,07
$\frac{a(b + c - a)}{ - 2} = \\ = \frac{0.02(-11.05 + 1.07 - 0.02)}{ - 2} = \\ = \frac{0.02 \times (-10)}{ - 2} = 0.01 \times 10 = \\ = 0.1$