Question

giải các phương trình

a)5+(96/x^2-16)=(2x-1/x+4)-(3x-1/4-x)

b)(3x+2/3x-2)-(6/2+3x)=9x^2/9x^2-4

c)(x+1/x^2+x+1)-(x-1/x^2-x+1)=3/x(x^4+x^2+1)

Answer 1

a) ĐKXĐ: \(x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)

<=> 20x² - 16x + 64 = 18x² + 8x

<=> 20x² - 16x + 64 - 18x² - 8x = 0

<=> 2x² - 24x + 64 = 0

<=> 2(x² - 12x + 32) = 0

<=> 2(x - 8)(x - 4) = 0

<=> (x - 8)(x - 4) = 0

<=> x - 8 = 0 hoặc x - 4 = 0

<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)

=> x = 8

b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)

\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)

<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

<=> (2 + 3x)² - 6(3x - 2) = 9x²

<=> 16 - 6x + 9x² = 9x²

<=> 16 - 6x + 9x² - 9x² = 0

<=> 16 - 6x = 0

<=> -6x = 0 - 16

<=> -6x = -16

<=> x = -16/-6 = 8/3

=> x = 8/3