Question : The value of $\frac{a^2-(b-c)^2}{(a+c)^2-b^2}+\frac{b^2-(a-c)^2}{(a+b)^2-c^2}+\frac{c^2-(a-b)^2}{(b+c)^2-a^2}$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 3
Correct Answer: 1
Solution : $\frac{a^2-(b-c)^2}{(a+c)^2-b^2}+\frac{b^2-(a-c)^2}{(a+b)^2-c^2}+\frac{c^2-(a-b)^2}{(b+c)^2-a^2}$ = $\frac{(a-b+c)(a+b-c)}{(a+c-b)(a+c+b)} + \frac{(b-a+c)(b+a-c)}{(a+b-c)(a+b+c)} + \frac{(c-a+b)(c+a-b)}{(b+c-a)(b+c+a)}$ = $\frac{(a+b-c)}{(a+c+b)} + \frac{(b-a+c)}{(a+b+c)} + \frac{(c+a-b)}{(b+c+a)}$ = $\frac{a+b-c+b-a+c+c+a-b}{a+b+c}$ = $\frac{a+b+c}{a+b+c}$ = $1$ Hence, the correct answer is 1.
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Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
Question : If $(a^2 = b + c)$, $(b^2 = a + c)$, $(c^2 = b + a)$. Then, what will be the value of $(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})$?
Question : If $a+b+c=0$, then the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}$ is:
Question : If $a, b, c$ are all non-zero and $a+b+c=0$, then find the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{ab}$.
Question : If $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}$, then $(x+y+z)$ is:
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