Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{41}$
Option 2: $\sqrt{29}$
Option 3: $\sqrt{23}$
Option 4: $\sqrt{43}$
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Correct Answer: $\sqrt{29}$
Solution : $x=\frac{1}{x-5}(x>0)$ $⇒x^2-5x-1=0$ $⇒x^2-1=5x$ Multiplying both sides by $\frac{1}{x}$, we get, $⇒x-\frac{1}{x}=5$ Squaring both sides, we get, $⇒(x-\frac{1}{x})^2=5^2$ $⇒x^2+\frac{1}{x^2}-2=25$ Adding 4 to both sides, we get, $⇒x^2+\frac{1}{x^2}+2=25+4$ $⇒(x+\frac{1}{x})^2=29$ $\therefore x+\frac{1}{x}=\sqrt{29}$ Hence, the correct answer is $\sqrt{29}$.
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Question : If $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$, and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right) ?$
Option 1: $22970 \sqrt{23}$
Option 2: $23030 \sqrt{23}$
Option 3: $23060 \sqrt{23}$
Option 4: $22960 \sqrt{23}$
Question : If $x^2+y^2=427$ and $xy=202$, then find the value of $\frac{x+y}{x-y}$.
Option 1: $\sqrt{\frac{835}{23}}$
Option 2: $\sqrt{\frac{830}{29}}$
Option 3: $\sqrt{\frac{831}{23}}$
Option 4: $\sqrt{\frac{830}{23}}$
Question : If $x^2-5\sqrt{5}x+1=0$ and $x > 0$, what is the value of $(x^3-\frac{1}{x^3})$?
Option 1: 1331
Option 2: 1364
Option 3: 1296
Option 4: 1244
Question : If $x^3=270+y^3$ and $x=(6+y)$, then what is the value of $(x+y)? $(given that $x>0$ and $y>0$)
Option 1: $2 \sqrt{3}$
Option 2: $\sqrt{3}$
Option 3: $4 \sqrt{3}$
Option 4: $4 \sqrt{2}$
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
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