If x is real,the maximum value of frac{3x^2 + | vedclass

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(C) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.Rearranging the terms,we get $y(3x^2 + 9x + 7) = 3x^2 + 9x + 17$.$3x^2(y - 1) + 9x(y - 1) + 7y - 17

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$41$

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(C) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.Rearranging the terms,we get $y(3x^2 + 9x + 7) = 3x^2 + 9x + 17$.$3x^2(y - 1) + 9x(y - 1) + 7y - 17 = 0$.Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$ $(D \geq 0)$.$D = [9(y - 1)]^2 - 4(3)(y - 1)(7y - 17) \geq 0$.$81(y - 1)^2 - 12(y - 1)(7y - 17) \geq 0$.Dividing by $3(y - 1)$ (assuming $y 
eq 1$): $27(y - 1) - 4(7y - 17) \geq 0$.$27y - 27 - 28y + 68 \geq 0$.$-y + 41 \geq 0 \Rightarrow y \leq 41$.Also,checking the case $y=1$: $3x^2 + 9x + 17 = 3x^2 + 9x + 7 \Rightarrow 17 = 7$,which is impossible. Thus $y 
eq 1$.By analyzing the range,the maximum value of $y$ is $41$.

If $x$ is real,the maximum value of $\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$ is

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