Chủ đề:
Bài 3: Phương trình đưa được về dạng ax + b = 0Câu hỏi:
1) 3 - 2x + 4 + 6x = x + 7 + 3x
2) -6(1,5 - 2x) = 3(-15 + 2x)
3) 3(2x - 5) + 5(x -1) = 4(x + 1)
1) \(\dfrac{4x+7}{x-1}\) = \(\dfrac{12x+5}{3x+4}\)
2) \(\dfrac{x}{x-1}\) - \(\dfrac{2x}{x^{2^{ }}-1}\) = 0
3) \(\dfrac{1}{3-x}\) - \(\dfrac{14}{x^2-9}\) = 1
4) \(\dfrac{x+1}{x-1}\) - \(\dfrac{x-1}{x+1}\) = \(\dfrac{4}{x^2-1}\)
5) x + \(\dfrac{1}{x}\) = x2 + \(\dfrac{1}{x^2}\)
6) \(\dfrac{x-1}{x^2+4}\) = \(\dfrac{x-1}{x+1}\)
1) \(\dfrac{7x-3}{x-1}\) = \(\dfrac{2}{3}\)
2) \(\dfrac{2\left(3-7x\right)}{1+x}\) = \(\dfrac{1}{2}\)
3) \(\dfrac{x^{2^{ }}-6}{x}\) = x + \(\dfrac{3}{2}\)
4) \(\dfrac{5}{3x+2}\) = 2x - 1
5) \(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}\) = 0
6) \(\dfrac{1}{x-2}\) + 3 = \(\dfrac{3-x}{x-2}\)