Question : The value of $x$ which satisfies the equation $\frac{x \:+\: a^2 \:+\: 2c^2}{b \:+\: c} + \frac{x \:+\: b^2 \:+\: 2a^2}{c+a} + \frac{x \:+\: c^2 \:+\: 2b^2}{a \:+\: b} = 0$ is:
Option 1: $(a^2+b^2+c^2)$
Option 2: $- (a^2+b^2+c^2)$
Option 3: $(a^2+2b^2+c^2)$
Option 4: $-(a^2+b^2+2c^2)$
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Correct Answer: $- (a^2+b^2+c^2)$
Solution : Given: $\frac{x+a^2+2c^2}{b+c}+\frac{x+b^2+2a^2}{c+a}+\frac{x+c^2+2b^2}{a+b}=0$ We know the algebraic identity, $(x^2-y^2)=(x+y)(x-y)$. Substitute the value of $x=-(a^2+b^2+c^2)$ in the given equation, we get, $\frac{–a^2–b^2–c^2+a^2+2c^2}{b+c}+\frac{–a^2–b^2–c^2+b^2+2a^2}{c+a}+\frac{–a^2–b^2–c^2+c^2+2b^2}{a+b}=0$ $⇒\frac{c^2–b^2}{b+c}+\frac{a^2–c^2}{c+a}+\frac{b^2–a^2}{a+b}=0$ $⇒\frac{(c–b)(b+c)}{b+c}+\frac{(a–c)(c+a)}{c+a}+\frac{(b–a)(a+b)}{a+b}=0$ $⇒c-b+a-c+b-a=0$ $⇒0=0$ It satisfies the equation. Hence, the correct answer is $-(a^2+b^2+c^2)$.
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Question : If $\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}=\frac{1}{2}$, then the value of $\frac{1}{1-2a}+\frac{1}{1-2b}+\frac{1}{1-2c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4
Question : If $x+\frac{1}{x}=c+\frac{1}{c}$, then the value of $x$ is:
Option 1: $c,\frac{1}{c}$
Option 2: $c,c^{2}$
Option 3: $c,2c$
Option 4: $0, 1$
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:
Option 1: 0
Option 2: 1
Option 4: 2
Question : If $a+b=2c$, then the value of $\frac{a}{a–c}+\frac{c}{b–c}$ is equal to (where $a\neq b\neq c$):
Option 1: $–1$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{2}$
Question : The graphs of the equations $4 x+\frac{1}{3} y=\frac{8}{3}$ and $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ intersect at a point P. The point P also lies on the graph of the equation:
Option 1: $x + 2y - 5 = 0$
Option 2: $3x - y - 7 = 0$
Option 3: $x - 3y - 12= 0$
Option 4: $4x - y + 7= 0$
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