Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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Correct Answer: $\frac{5}{2}$
Solution : Given: $x+\frac{1}{x}=3$ To find: $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ Dividing with $x$ on both numerator and denominator, = $\frac{3x-4+\frac{3}{x}}{x-1+\frac{1}{x}}$ = $\frac{3(x+\frac{1}{x})-4}{(x+\frac{1}{x})-1}$ = $\frac{(3×3)-4}{3-1}$ = $\frac{5}{2}$ Hence, the correct answer is $\frac{5}{2}$.
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Question : If $x+\frac{1}{x}=2$, then find the value of $x^5+\frac{1}{x^5}$.
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 2
Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Option 1: $\frac{3}{7}$
Option 2: $1\frac{1}{7}$
Option 3: $1$
Option 4: $2$
Question : If $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Option 1: $\frac{5}{7}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{2}{7}$
Question : If $4(x+5)-3>6-4x\geq x-5$, then the value of $x$ is:
Option 1: –2
Option 2: –1
Option 3: 3
Option 4: 4
Question : If $2x+\frac{1}{4x}=1$, then the value of $x^{2}+\frac{1}{64x^{2}}$ is:
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{4}$
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