Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4
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Correct Answer: 1
Solution : Given: $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$ So, $a^2=b+c$, $b^2=c+a$ and $c^2=a+b$. Now, $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ Multiplying the numerator and denominator of the first, second and third term by $a$, $b$, and $c$, respectively. = $\frac{a}{a+a^2}+\frac{b}{b+b^2}+\frac{c}{c+c^2}$ = $\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}$ (putting the values of $a^2$, $b^2$ and $c^2$) = $\frac{a+b+c}{a+b+c}$ = 1 Hence, the correct answer is 1.
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Question : If $a+b+c = 0$, then the value of $\small \frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$ is:
Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$.
Question : If $a^{2}=b+c,b^{2}=c+a,c^{2}=a+b$, the value of $3\left (\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$ is:
Question : If $\frac{a^{2} - bc}{a^{2}+bc}+\frac{b^{2}-ca}{b^{2}+ca}+\frac{c^{2}-ab}{c^{2}+ab}=1$, then the value of $\frac{a^{2}}{a^{2}+bc}+\frac{b^{2}}{b^{2}+ac}+\frac{c^{2}}{c^{2}+ab}$ is:
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ where $a \neq b\neq c\neq 0$, then the value of $a^{2}b^{2}c^{2}$ is:
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