Question : If $a,b,c$, and $d$ satisfy the equations: $a+7b+3c+5d=0$, $8a+4b+6c+2d=–4$, $2a+6b+4c+8d=4$, $5a+3b+7c+d=–4$, then $\frac{a+d}{b+c}$?
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: – 4
Correct Answer: –1
Solution : Given: $8a+4b+6c+2d=–4$ (equation 1) $2a+6b+4c+8d=4$ (equation 2) Adding equations 1 and 2, we get, $8a+4b+6c+2d+2a+6b+4c+8d=–4+4$ ⇒ $10(a+b+c+d)=0$ ⇒ $(a+d)=–(b+c)$ ⇒ $\frac{a+d}{b+c}$ = –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: $(a\neq b\neq c)$
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $3$
Question : If $x=\frac{8ab}{a+b}(a\neq b),$ then the value of $\frac{x+4a}{x–4a}+\frac{x+4b}{x–4b}$ is:
Option 3: 2
Option 4: 4
Question : If $a+b=2c$, then find $\frac{a}{a–c}+\frac{c}{b–c}$:
Option 4: –1
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:
Option 1: $a+b+c$
Option 2: $0$
Option 3: $a^2+b^2+c^2$
Option 4: $2$
Question : If $\left (2a-3 \right )^{2}+\left (3b+4 \right )^{2}+\left ( 6c+1\right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$ is:
Option 1: $abc+3$
Option 2: $6$
Option 3: $0$
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