2x+13x=5,2x+13x=5,2x + \frac{1}{{3x}} = 5{\text{,}} then the value of 5x6x2+20x+1" role="presentation">5x6x2+20x+15x6x2+20x+1\frac{{5x}}{{6{x^2} + 20x + 1}} is?" /> 2x+13x=5,2x+13x=5,2x + \frac{1}{{3x}} = 5{\text{,}} then the value of 5x6x2+20x+1" role="presentation">5x6x2+20x+15x6x2+20x+1\frac{{5x}}{{6{x^2} + 20x + 1}} is?" /> 2x+13x=5,2x+13x=5,2x + \frac{1}{{3x}} = 5{\text{,}} then the value of 5x6x2+20x+1" role="presentation">5x6x2+20x+15x6x2+20x+1\frac{{5x}}{{6{x^2} + 20x + 1}} is?" />

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If $2 x + \frac{1}{3 x} = 5,$ $2x + \frac{1}{{3x}} = 5{\text{,}}$ then the value of $\frac{5 x}{6 x^{2} + 20 x + 1}$ $\frac{{5x}}{{6{x^2} + 20x + 1}}$ is?

1 $\frac{1}{4}$ $\frac{1}{4}$
2 $\frac{1}{6}$ $\frac{1}{6}$
3 $\frac{1}{5}$ $\frac{1}{5}$
4 $\frac{1}{7}$ $\frac{1}{7}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 2 x + \frac{1}{3 x} = 5 \\ 6 x^{2} + 1 = 15x . . . . . . (i) \\ Now, \frac{5 x}{6 x^{2} + 20 x + 1} \\ = \frac{5 x}{6 x^{2} + 1 + 20 x} \\ [From equation (i)] \\ = \frac{5 x}{15 x + 20 x} \\ = \frac{5 x}{35 x} \\ = \frac{1}{7} \end{aligned}

$\eqalign{ & 2x + \frac{1}{{3x}} = 5 \cr & 6{x^2}{\text{ + 1 = 15x}}\,......{\text{(i)}} \cr & {\text{Now,}}\frac{{5x}}{{6{x^2} + 20x + 1}} \cr & = \frac{{5x}}{{6{x^2} + 1 + 20x}} \cr & \left[ {{\text{From equation (i)}}} \right] \cr & = \frac{{5x}}{{15x + 20x}}{\text{ }} \cr & = \frac{{5x}}{{35x}} \cr & = \frac{1}{7} \cr}$

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then the value of

\frac{5 x}{6 x^{2} + 20 x + 1}

$\frac{{5x}}{{6{x^2} + 20x + 1}}$ is?", "text": "If

2 x + \frac{1}{3 x} = 5,

$2x + \frac{1}{{3x}} = 5{\text{,}}$ then the value of

\frac{5 x}{6 x^{2} + 20 x + 1}

$\frac{{5x}}{{6{x^2} + 20x + 1}}$ is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 2 x + \frac{1}{3 x} = 5 \\ 6 x^{2} + 1 = 15x . . . . . . (i) \\ Now, \frac{5 x}{6 x^{2} + 20 x + 1} \\ = \frac{5 x}{6 x^{2} + 1 + 20 x} \\ [From equation (i)] \\ = \frac{5 x}{15 x + 20 x} \\ = \frac{5 x}{35 x} \\ = \frac{1}{7} \end{aligned}

$\eqalign{ & 2x + \frac{1}{{3x}} = 5 \cr & 6{x^2}{\text{ + 1 = 15x}}\,......{\text{(i)}} \cr & {\text{Now,}}\frac{{5x}}{{6{x^2} + 20x + 1}} \cr & = \frac{{5x}}{{6{x^2} + 1 + 20x}} \cr & \left[ {{\text{From equation (i)}}} \right] \cr & = \frac{{5x}}{{15x + 20x}}{\text{ }} \cr & = \frac{{5x}}{{35x}} \cr & = \frac{1}{7} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 2 x + \frac{1}{3 x} = 5 \\ 6 x^{2} + 1 = 15x . . . . . . (i) \\ Now, \frac{5 x}{6 x^{2} + 20 x + 1} \\ = \frac{5 x}{6 x^{2} + 1 + 20 x} \\ [From equation (i)] \\ = \frac{5 x}{15 x + 20 x} \\ = \frac{5 x}{35 x} \\ = \frac{1}{7} \end{aligned}

$\eqalign{ & 2x + \frac{1}{{3x}} = 5 \cr & 6{x^2}{\text{ + 1 = 15x}}\,......{\text{(i)}} \cr & {\text{Now,}}\frac{{5x}}{{6{x^2} + 20x + 1}} \cr & = \frac{{5x}}{{6{x^2} + 1 + 20x}} \cr & \left[ {{\text{From equation (i)}}} \right] \cr & = \frac{{5x}}{{15x + 20x}}{\text{ }} \cr & = \frac{{5x}}{{35x}} \cr & = \frac{1}{7} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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