Question

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[tex] \frac{18}{ {x}^{2} + 3x} - \frac{4}{x} + \frac{6}{x + 3} = \dots[/tex]
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Selamat Berjuang​

Answer 1

Verified answer

[tex]\frac{18}{x^{2} +3x}-\frac{4}{x} +\frac{6}{x+3}\\\\\frac{18}{x(x+3)}-\frac{4}{x} +\frac{6}{x+3} \\\frac{18}{x(x+3)}- \frac{4(x+3)}{x(x+3)} - \frac{6x}{x(x+3)} \\\frac{18-4(x+3)-6x}{x(x+3)} \\\frac{2x+6}{x(x+3)}\\\frac{2(x+3)}{x(x+3)} = \frac{2}{x}[/tex]

Answer 2

Jawab:

Jawaban paling sederhana dari persamaan di atas adalah 2/x.

Penjelasan dengan langkah-langkah:

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{18}{x^2+3x} - \frac{4(x+3)}{x(x+3)} + \frac{6(x)}{(x+3)(x)}[/tex]

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{18}{x^2+3x} - \frac{4x + 12}{x^2+3x} + \frac{6x}{x^2+3x}[/tex]

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{18-(4x+12)+6x}{x^2+3x}[/tex]

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{18 - 4x - 12 + 6x}{x^2+3x}[/tex]

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{6 + 2x}{x^2+3x}[/tex]

[tex]\frac{18}{x^2+3x} - \frac{4}{x} + \frac{6}{x+3}[/tex] = [tex]\frac{2(x+3)}{x(x+3)}[/tex] = 2/x

Semoga membantu!!