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Answer :
The answer is C.2x-3/x,(x_0)
[tex] \frac{10 x^{2} -15x}{5 x^{2} } = \frac{2x*5x -3*5x}{5 x^{2} }= \frac{5x(2x-3)}{5x*x} [/tex]
5x can be canceled out, thus:
[tex]\frac{5x(2x-3)}{5x*x} = \frac{2x-3}{x} [/tex]
Since x is denominator and if denominator is 0, the terms would be undefined, then x≠ 0
[tex] \frac{10 x^{2} -15x}{5 x^{2} } = \frac{2x*5x -3*5x}{5 x^{2} }= \frac{5x(2x-3)}{5x*x} [/tex]
5x can be canceled out, thus:
[tex]\frac{5x(2x-3)}{5x*x} = \frac{2x-3}{x} [/tex]
Since x is denominator and if denominator is 0, the terms would be undefined, then x≠ 0
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