Question

1) Tìm x:

a. \(x^3-\dfrac{1}{4}x=0\) b. \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

c. \(2x^2-x-6=0\)

2) Cho biểu thức \(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\)

a.Rút gọn B.

b. Tính giá trị của B tại x = 2

Answer 1

a) \(x^3-\dfrac{1}{4}x=0\)

⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)

⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)

b) (2x - 1)² - (x + 3)² = 0

⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0

⇔ (x - 4)(3x +2) = 0

⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)

c) 2x² - x - 6 = 0

⇔ 2x² - 4x + 3x - 6 = 0

⇔ 2x(x - 2) + 3(x - 2) = 0

⇔ (x - 2) (2x + 3) = 0

⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)

Answer 2

2)a.

\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)

b)

\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)