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Anie S.

Calculus 1 / AB

6 days, 5 hours ago

$$ \frac{1}{x^{2}-3 x+3}+\frac{2}{x^{2}-3 x+4}=\frac{6}{x^{2}-3 x+5} $$

Chapter 7

Equations And Inequalities

in the struggle. Let us take Why? As X squared minus three X. Which makes the given equation one upon Y plus three plus two upon y plus four is equals to six upon Y plus five. Taking L. C. You get Y plus four plus To buy plus six Upon y plus three In two y plus four is equals to six upon Y plus five. This is nothing but three. Right Plus then In do I plus five Is equal to six and 2. Y plus three. And to y plus four. Opening this disk comes as three Us square plus 25. Y plus 50 is equals to 6. 5 square plus 42. Y plus 72. Moving the terms around we get three by square plus 17. Y plus 22 equals to zero. This effect raised us Y plus three and do three y plus 11 Is equals to zero as we get biased -3 Or -11 x three. So let us take case one where y plus three zero which means that x squared minus three X Plus three is equals to zero. I know that this is always greater than zero. Hence this is never possible. Second cases three bypass 11 0 which is three X squared minus nine x plus 11 is equals to zero solving this. We get access nine plus minus nine squared is 81 4312 and 2 11 by six. I know that the the determinant value sorry discriminative value here which is this value Is next to zero. Hence there is no solution is this is always.

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