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Sagot :
To solve the problem, we start with the equation:
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
We need to find the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex].
First, solve [tex]\( x + \frac{1}{x} = 3 \)[/tex]. Rewrite the equation to form a quadratic equation:
[tex]\[ x + \frac{1}{x} = 3 \implies x^2 + 1 = 3x \implies x^2 - 3x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\( x^2 - 3x + 1 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex].
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus, the two solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{3 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{3 + \sqrt{5}}{2} \][/tex]
Next, we need to determine the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex] for each solution.
For [tex]\( x_1 = \frac{3 - \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_1^2 = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_1^2 + 1 = \frac{7 - 3\sqrt{5}}{2} + 1 = \frac{7 - 3\sqrt{5} + 2}{2} = \frac{9 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_1}{x_1^2 + 1} = \frac{6 \left(\frac{3 - \sqrt{5}}{2}\right)}{\frac{9 - 3\sqrt{5}}{2}} = \frac{6 (3 - \sqrt{5})}{9 - 3\sqrt{5}} = \frac{18 - 6\sqrt{5}}{9 - 3\sqrt{5}} = \frac{6 (3 - \sqrt{5})}{3 (3 - \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_1}{x_1^2 + 1} = 2 \][/tex]
For [tex]\( x_2 = \frac{3 + \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_2^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2 = \frac{(3 + \sqrt{5})^2}{4} = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_2^2 + 1 = \frac{7 + 3\sqrt{5}}{2} + 1 = \frac{7 + 3\sqrt{5} + 2}{2} = \frac{9 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{6 \left(\frac{3 + \sqrt{5}}{2}\right)}{\frac{9 + 3\sqrt{5}}{2}} = \frac{6 (3 + \sqrt{5})}{9 + 3\sqrt{5}} = \frac{18 + 6\sqrt{5}}{9 + 3\sqrt{5}} = \frac{6 (3 + \sqrt{5})}{3 (3 + \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = 2 \][/tex]
Although through careful simplification, it was found that both results amount to 2, it turns out the correct calculated was:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{3\sqrt{5} + 9}{1 + ( \sqrt{5}/2 + 3/2) ^2} \][/tex]
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
We need to find the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex].
First, solve [tex]\( x + \frac{1}{x} = 3 \)[/tex]. Rewrite the equation to form a quadratic equation:
[tex]\[ x + \frac{1}{x} = 3 \implies x^2 + 1 = 3x \implies x^2 - 3x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\( x^2 - 3x + 1 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex].
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus, the two solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{3 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{3 + \sqrt{5}}{2} \][/tex]
Next, we need to determine the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex] for each solution.
For [tex]\( x_1 = \frac{3 - \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_1^2 = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_1^2 + 1 = \frac{7 - 3\sqrt{5}}{2} + 1 = \frac{7 - 3\sqrt{5} + 2}{2} = \frac{9 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_1}{x_1^2 + 1} = \frac{6 \left(\frac{3 - \sqrt{5}}{2}\right)}{\frac{9 - 3\sqrt{5}}{2}} = \frac{6 (3 - \sqrt{5})}{9 - 3\sqrt{5}} = \frac{18 - 6\sqrt{5}}{9 - 3\sqrt{5}} = \frac{6 (3 - \sqrt{5})}{3 (3 - \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_1}{x_1^2 + 1} = 2 \][/tex]
For [tex]\( x_2 = \frac{3 + \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_2^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2 = \frac{(3 + \sqrt{5})^2}{4} = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_2^2 + 1 = \frac{7 + 3\sqrt{5}}{2} + 1 = \frac{7 + 3\sqrt{5} + 2}{2} = \frac{9 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{6 \left(\frac{3 + \sqrt{5}}{2}\right)}{\frac{9 + 3\sqrt{5}}{2}} = \frac{6 (3 + \sqrt{5})}{9 + 3\sqrt{5}} = \frac{18 + 6\sqrt{5}}{9 + 3\sqrt{5}} = \frac{6 (3 + \sqrt{5})}{3 (3 + \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = 2 \][/tex]
Although through careful simplification, it was found that both results amount to 2, it turns out the correct calculated was:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{3\sqrt{5} + 9}{1 + ( \sqrt{5}/2 + 3/2) ^2} \][/tex]
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