Đáp án:
Giải thích các bước giải:
`a, (2x-1)/(x^2-1) = 6/(3x+1)`
ĐKXĐ : \(\left\{\begin{array}{l}x\ne1\\x\ne-1\\x\ne-\dfrac{1}{3}\end{array} \right.\)
`-> (2x-1)(3x+1) = (x^2-1)*6`
`-> 6x^2 - x - 1 = 6x^2 - 6`
`-> -x - 1 = -6`
`-> -x = -5`
`-> x = 5(TM)`
Vậy `x \in {5}`
`b, 1/(x+3) + 1/(x-1) = 10/((x+3)(x-1))`
`-> (1(x-1))/((x+3)(x-1)) + (1(x+3))/((x+3)(x-1)) = 10/((x+3)(x-1))`
`-> (x-1) + (x+3) = 10`
`-> (x+x) - 1 + 3 = 10`
`-> 2x + 2 = 10`
`-> 2x = 8`
`-> x = 4(TM)`
Vậy `x \in {4}`
$\\$
`a, 1/(x+1) - 4/(x^2-x+1) = (2x^2+1)/(x^3+1)`
ĐKXĐ : `x \ne -1`
`-> (1(x^2-x+1))/((x+1)(x^2-x+1)) - (4(x+1))/((x+1)(x^2-x+1)) = (2x^2+1)/((x+1)(x^2-x+1))`
`-> x^2 - x + 1 - 4(x+1) = 2x^2 + 1`
`-> x^2 - 5x - 3 = 2x^2 + 1`
`-> x^2 - 5x - 4 = 2x^2`
`-> -x^2 - 5x - 4 = 0`
`-> -(x^2+5x+4) = 0`
`-> -(x+1)(x+4) = 0`
`->`\(\left\{ \begin{array}{l}x+1=0\\x+4=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-1(KTM)\\x=-4(TM)\end{array} \right.\)
Vậy `x \in {-4}`
`b, (3x+1)/(x+1) - (2x-5)/(x-3) = 1 - 4/((x+1)(x-3))`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne-1\\x\ne3\end{array} \right.\)
`-> ((3x+1)(x-3))/((x+1)(x-3)) - ((2x-5)(x+1))/((x+1)(x-3)) = 1(x+1)(x-3) - 4/((x+1)(x-3))`
`-> (3x+1)(x-3) - (2x-5)(x+1) = (x+1)(x-3) - 4`
`-> x^2 - 5x + 2 = x^2 - 2x - 7`
`-> x^2 - 5x = x^2 - 2x - 9`
`-> -5x = -2x - 9`
`-> -3x = -9`
`-> x = 3(KTM)`
Vậy `x \in ∅`
