Question

câu 47: tìm giá trị của ( x ) thỏa mãn phương trình (\frac{x}{2x - 6} + \frac{x}{2x + 2} = \frac{-2x}{3 - x x + 1}) (note: there may be a formatting error in the original equations right - hand side denominator, assuming it should be a single denominator or a typo. the main idea is to solve the rational equation for ( x ))

Explanation:

Step1: 化简分母

先对原方程的分母进行因式分解：\(2x - 6 = 2(x - 3)\)，\(2x + 2 = 2(x + 1)\)，原方程变为\(\frac{x}{2(x - 3)} + \frac{x}{2(x + 1)} = \frac{-2x}{3 - x} + \frac{-2x}{x + 1}\)。注意到\(\frac{-2x}{3 - x}=\frac{2x}{x - 3}\)，所以方程进一步化为\(\frac{x}{2(x - 3)} + \frac{x}{2(x + 1)} = \frac{2x}{x - 3} - \frac{2x}{x + 1}\)。

Step2: 去分母

方程两边同时乘以\(2(x - 3)(x + 1)\)（\(x
eq3\)且\(x
eq - 1\)）去分母得：\(x(x + 1)+x(x - 3)=4x(x + 1)-4x(x - 3)\)。

Step3: 展开并整理方程

展开左边：\(x^{2}+x+x^{2}-3x = 2x^{2}-2x\)；展开右边：\(4x^{2}+4x - 4x^{2}+12x = 16x\)。此时方程变为\(2x^{2}-2x = 16x\)。

Step4: 移项求解

移项得\(2x^{2}-2x - 16x = 0\)，即\(2x^{2}-18x = 0\)，提取公因式\(2x\)得\(2x(x - 9)=0\)。

Step5: 求解方程

由\(2x(x - 9)=0\)，根据“若\(ab = 0\)，则\(a = 0\)或\(b = 0\)”，可得\(2x = 0\)或\(x - 9 = 0\)，解得\(x = 0\)或\(x = 9\)。

Step6: 检验根

• 当\(x = 0\)时，原方程分母\(2x - 6=-6

eq0\)，\(2x + 2 = 2
eq0\)，\(3 - x = 3
eq0\)，\(x + 1 = 1
eq0\)，所以\(x = 0\)是原方程的根。

• 当\(x = 9\)时，原方程分母\(2x - 6 = 12

eq0\)，\(2x + 2 = 20
eq0\)，\(3 - x=-6
eq0\)，\(x + 1 = 10
eq0\)，所以\(x = 9\)是原方程的根。

Answer:

\(x = 0\)或\(x = 9\)