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mezya [45]
3 years ago
14

X+2/x-3 + x-3/x+2 = 5/2

Mathematics
1 answer:
White raven [17]3 years ago
8 0

Answer:

x_{1} =\frac{-17- \sqrt{129}}{2},  x_{2} =\frac{-17+ \sqrt{129}}{2}

Step-by-step explanation:

\frac{x + 2}{x + 3}  +  \frac{x - 3}{x + 2}  =  \frac{5}{2}

\frac{x + 2}{x + 3}  +  \frac{x - 3}{x + 2}  =  \frac{5}{2}

\frac{x + 2}{x + 3}  +  \frac{x - 3}{x + 2}  -  \frac{5}{2}  = 0

\frac{2(x + 2 {)}^{2} + 2(x + 3) \times (x - 3) - 5(x + 3) \times (x + 2) }{2(x + 3) \times (x + 2)}  = 0

\frac{2(x + 2 {)}^{2} + 2( {x}^{2} - 9) + ( - 15x - 15) \times (x + 2)  }{2(x + 3) \times (x + 2)} = 0

\frac{2( {x}^{2}  + 4x + 4) - 3 {x}^{2}  - 48 - 25x}{2(x + 3) \times (x + 2)} = 0

\frac{2 {x}^{2} + 8x + 8 - 3 {x}^{2} - 48 - 25x  }{2(x + 3) \times (x + 2)}  = 0

\frac{ -  {x}^{2} - 17x - 40 }{2(x + 3) \times (x + 2)}  = 0

{x}^{2}  + 17x + 40 = 0

x =  \frac{ - 17± \sqrt{ {17}^{2} - 4 \times 1 \times 40 } }{2 \times 1}

x =  \frac{ - 17± \sqrt{289 - 160} }{2}

x =  \frac{ - 17± \sqrt{129} }{2}

x_{1} =\frac{-17- \sqrt{129}}{2},  x_{2} =\frac{-17+ \sqrt{129}}{2}

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The correct statements about the equation are b, c, and e.

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We have to determine

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brainly.com/question/17822016

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