Please solve q 10?

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Answer 1 · 2018-04-01T18:02:51

The answer is $= 2$ $=2$

$x = \frac{a}{b + c}$ $x=a/(b+c)$

$y = \frac{b}{c + a}$ $y=b/(c+a)$

$z = \frac{c}{a + b}$ $z=c/(a+b)$

Therefore,

$\frac{1}{1 + x} = \frac{1}{1 + \frac{a}{b + c}} = \frac{b + c}{a + b + c}$ $1/(1+x)=1/(1+a/(b+c))=(b+c)/(a+b+c)$

$\frac{1}{1 + y} = \frac{1}{1 + \frac{b}{c + a}} = \frac{c + a}{a + b + c}$ $1/(1+y)=1/(1+b/(c+a))=(c+a)/(a+b+c)$

$\frac{1}{1 + z} = \frac{1}{1 + \frac{c}{a + b}} = \frac{a + b}{a + b + c}$ $1/(1+z)=1/(1+c/(a+b))=(a+b)/(a+b+c)$

Finally,

$\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z}$ $1/(1+x)+1/(1+y)+1/(1+z)$

$= \frac{b + c}{a + b + c} + \frac{c + a}{a + b + c} + \frac{a + b}{a + b + c}$ $=(b+c)/(a+b+c)+(c+a)/(a+b+c)+(a+b)/(a+b+c)$

$= \frac{b + c + c + a + a + b}{a + b + c}$ $=(b+c+c+a+a+b)/(a+b+c)$

$= \frac{2 (a + b + c)}{a + b + c}$ $=(2(a+b+c))/(a+b+c)$

$= 2$ $=2$