Tìm x, biết: 2/x + 3 + 3x^2 - 9 = 0 ( x khác 3) A. x = 0 B. x = 1/2 C. x = 1 D. x = 3/2

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Tìm x, biết: \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}} = 0\,\,\,\left( {{\rm{x}} \ne \pm \,3} \right)\]

x = 0

\(x = \frac{1}{2}\)

x = 1

\[{\rm{x}} = \frac{3}{2}\]

Giải thích

Lời giải

Đáp án đúng là: D

Ta có \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}}\]\[ = \frac{2}{{{\rm{x}} + 3}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

\[ = \frac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

\[ = \frac{{2\left( {x - 3} \right) + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 6 + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

Mà \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}} = 0\]nên \[\frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\]

\[2x - 3 = 0\]

\[2x = 3\]

\[x = \frac{3}{2}\]

Vậy \[x = \frac{3}{2}\].