If $$x + \frac{1}{x} = 3{\text{,}}$$ then the value of $$\frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}}$$ is?
A. $$\frac{4}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{2}$$
D. $$\frac{5}{3}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr & = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr & = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr & = \frac{{9 - 4}}{2} \cr & = \frac{5}{2} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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