Question

a) \(\frac{1}{x-4}-\frac{2+x}{x+4}=\frac{x\left(7-x\right)}{x^2-16}\)

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

Answer 1

a, \(\frac{1}{x-4}-\frac{2+x}{x+4}=\frac{x.\left(7-x\right)}{x^2-16}\)

= \(\frac{x+4}{x^2-16}-\frac{\left(2+x\right)\left(x-4\right)}{x^2-16}=\frac{x.\left(7-x\right)}{x^2-16}\)

= \(x+4-2x+8-x^2+4x=7x-x^2\)

= x - 2x + 4x - 7x -xx + xx = -4 - 8

=> -4x = -12

x = 3

vậy phương trình có s = 3

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

= \(\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4.\left(x-1\right)}{x^3-1}\)

= \(x^2+x+1+2x^2-5=4x-4\)

= \(\left(x^2+2x^2\right)+x-4x=-1+5-4\)

= \(3x^2\) - 3x = 0

= 3x.(x - 1) = 0

=> 3x = 0 hoặc x-1=0

=> x = 0 x = 1