Question

help me bài này nhen

(\(\frac{3x}{x+3}\)+\(\frac{2}{x-5}\)):\(\frac{1}{x^2-2x-15}\)

help mee đi , dò xem kết quả đk ấy mà

Answer 1

⇔ \((\frac{3x}{x+3}+\frac{2}{x-5}):\frac{1}{\left(x-5\right)\left(x+3\right)}\)

ĐK : x \(\ne-3,\) x \(\ne5\)

\(\Leftrightarrow\left[\frac{3x\left(x-5\right)}{\left(x+3\right)\left(x-5\right)}+\frac{2\left(x+3\right)}{\left(x-5\right)\left(x+3\right)}\right]:\frac{1}{\left(x+3\right)\left(x-5\right)}\)

\(\Leftrightarrow\left[\frac{3x^2-15x+2x+6}{\left(\right)\left(\right)}\right]:\frac{1}{\left(\right)\left(\right)}\)

\(\Leftrightarrow\left[\frac{3x^2-13x+6}{\left(x-5\right)\left(x+3\right)}\right].\left(x+3\right)\left(x-5\right)\)

\(\Leftrightarrow3x^2-13x+6\)