\(\dfrac{1}{x^2+x} + \dfrac{1}{x^2+3x+2} = -2\)

Question

\(\dfrac{1}{x^2+x} + \dfrac{1}{x^2+3x+2} = -2\)

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Josephine 4 ngày 2021-09-15T10:20:31+00:00 1 Answers 0 views 0

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    2021-09-15T10:22:25+00:00

    \(DKXD:\begin{cases}x^2+x\ne 0\\x^2+3x+2\ne 0\end{cases}\\↔\begin{cases}x(x+1)\ne 0\\(x+1)(x+2)\ne 0\end{cases}↔\begin{cases}x\ne 0\\x+1\ne 0\\x+2\ne 0\end{cases}\\↔\begin{cases}x\ne 0\\x\ne -1\\x\ne -2\end{cases}\\\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}=-2\\↔\dfrac{1}{x(x+1)}+\dfrac{1}{x^2+2x+x+2}=-2\\↔\dfrac{1}{x(x+1)}+\dfrac{1}{x(x+2)+(x+2)}=-2\\↔\dfrac{1}{x(x+1)}+\dfrac{1}{(x+1)(x+2)}=-2\\↔\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+2=0\\↔\dfrac{1}{x}-\dfrac{1}{x+2}+2=0\\↔x+2-x+2x(x+2)=0\\↔2+2x^2+4x=0\\↔x^2+2x+1=0\\↔(x+1)^2=0\\↔x+1=0\\↔x=-1(KTM)\\\text{Vậy pt vô nghiệm}\,\, S=\varnothing\)

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