Đáp án:
$\\$
Giải thích các bước giải:
`a, (x+1)/x = (x^2+1)/x^2`
ĐKXĐ : `x \ne 0`
`-> (x+1)x^2 = x(x^2+1)`
`-> (x+1)x^2- x(x^2+1) = 0`
`-> x[x(x+1)-(x^2+1)] = 0`
`-> x(x-1) = 0`
`->`\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=0(KTM)\\x=1(TM)\end{array} \right.\)
Vậy `x \in {1}`
`b, (3x-2)/(x+7) = (6x+1)/(2x-3)`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne-7\\x\ne\dfrac{3}{2}\end{array} \right.\)
`-> (3x-2)(2x-3) = (x+7)(6x+1)`
`-> 6x^2 - 13x + 6 = 6x^2 + 43x + 7`
`-> 6x^2 - 13x = 6x^2 + 43x + 1`
`-> -13x = 43x + 1`
`-> -56x = 1`
`-> x = -1/56(TM)`
Vậy `x \in {-1/56}`
`c, (2x-2x^2)/(x+3) = (4x)/(x+3) + 2/7`
ĐKXĐ : `x \ne -3`
`-> (7(2x-2x^2))/(7(x+3)) = (7*4x)/(7(x+3)) + (2(x+3))/(7(x+3))`
`-> 7(2x-2x^2) = 28x - 2(x+3)`
`-> 14x - 14x^2 = 30x + 6`
`-> 14x - 14x^2 - 6 = 30x`
`-> -14x^2 - 16x - 6 = 0`
Do `-14x^2 - 16x - 6 < 0 ∀x`
`->` vô nghiệm
Vậy `x \in ∅`
