Question

Cho biểu thức H=\(1-\frac{1}{x^2-4}\left[\left(1-\frac{x^2+4}{4x}\right):\left(\frac{1}{x}-\frac{1}{2}\right)\right]\)

a)Tìm ĐKXĐ, rút gọn

b)Tìm x để H=\(\frac{1}{4}\)

Answer 1

a) ĐKXĐ : x\(\ne\)2,-2,0

H = 1 - \(\frac{1}{x^2-4}\)[(\(\frac{4x-x^2-4}{4x}\)) : (\(\frac{2-x}{2x}\))

H = 1 - \(\frac{1}{x^2-4}\)[\(\frac{-\left(x-2\right)^2}{4x}\).\(\frac{2x}{-\left(x-2\right)}\)]

H = 1 - \(\frac{1}{x^2-4}\).\(\frac{x-2}{2}\)

H = 1 - \(\frac{1}{2\left(x+2\right)}\)

H = \(\frac{2\left(x+2\right)-1}{2\left(x+2\right)}\)

H = \(\frac{2x+4-1}{2x+4}\)

H = \(\frac{2x+3}{2x+4}\)

b) H = \(\frac{1}{4}\)

⇒ \(\frac{2x+3}{2x+4}\)=\(\frac{1}{4}\) ĐKXĐ:x ≠ -2

⇒ 4(2x+3) = 2x+4

⇔ 8x+12 = 2x+4

⇔ 8x - 2x = 4 - 12

⇔ 6x = 8

⇔ x = \(\frac{4}{3}\) (TMĐKXĐ)

Vậy để H = \(\frac{1}{4}\) thì x = \(\frac{4}{3}\)