If $$a^2=by+cz,b^2=cz+ax,c^2=ax+by$$, then the value of $$\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$$ is
Given : $$a^2=by+cz$$
Adding $$'ax'$$ on both sides, we get :
=> $$a^2+ax=ax+by+cz$$
=> $$a(a+x)=ax+by+cz$$
=> $$\frac{1}{a+x}=\frac{a}{ax+by+cz}$$
=> $$\frac{x}{a+x}=\frac{ax}{ax+by+cz}$$ ----------(i)
Similarly, $$\frac{y}{b+y}=\frac{by}{ax+by+cz}$$ ----------(ii)
and $$\frac{z}{c+z}=\frac{cz}{ax+by+cz}$$ ----------(iii)
Adding equations (i),(ii) and (iii)
=>Â $$\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$$ $$= \frac{ax}{ax+by+cz}+\frac{by}{ax+by+cz}+\frac{cz}{ax+by+cz}$$
= $$\frac{ax+by+cz}{ax+by+cz}=1$$
=> Ans - (A)
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