Question : If $\{(1÷x)×(x^3-2x^2)+(4^3÷2^5)×x\}÷25=1$, then the possible values of $x$ is/are:
Option 1: $-5$
Option 2: $5$
Option 3: $0$
Option 4: Both $5$ and $-5$
Correct Answer: Both $5$ and $-5$
Solution : $\{(1÷x)×(x^3-2x^2)+(4^3÷2^5)×x\}÷25=1$ ⇒ $(x^2-2x)+2x=25$ ⇒ $x^2=25$ $\therefore x=\pm5$ Hence, the correct answer is both $5$ and $–5$.
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Question : If $x^4+x^{-4}=47, x>0$, then what is the value of $x+\frac{1}{x}-2?$
Option 1: 1
Option 2: 0
Option 3: 5
Option 4: 3
Question : If $x^2-5 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 21
Option 2: 22
Option 3: 25
Option 4: 24
Question : If $x^2-3 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 5
Option 2: 6
Option 3: 9
Option 4: 7
Question : If $x(x-5)=-1$, then the value of $x^3\left(x^3-110\right)=$?
Option 1: 0
Option 2: –1
Option 3: 1
Option 4: 2
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
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