Đáp án:
Giải thích các bước giải:
`a, 3(5x-2)-7x = 10`
`-> 15x - 6 - 7x = 10`
`-> 8x - 6 = 10`
`-> 8x = 16`
`-> x = 2`
Vậy `x \in {2}`
`b, (2x+1)/3 + (3x-2)/2 = 1/6`
`-> (2(2x+1))/6 + (3(3x-2))/6 = 1/6`
`-> 2(2x+1) + 3(3x-2) = 1`
`-> 4x + 2 + 9x - 3 = 1`
`-> 13x - 1 = 1`
`-> 13x = 2`
`-> x = 2/13`
Vậy `x \in {2/13}`
`c, x^3 - 3x^2 + 2x - 6 = 0`
`-> (x^3-3x^2) + (2x-6) = 0`
`-> 2(x-3) + x^2(x-3) = 0`
`-> (x-3)(x^2+2) = 0`
Vì `x^2 + 2 = 0 => x^2 = -2 => x ∈ ∅`
`-> x - 3 = 0`
`-> x = 3`
Vậy `x \in {3}`
`(x-1)/x + 1/(x+1) = (2x+1)/(x^2+x)`
ĐKXĐ : `x \ne -1 , x \ne 0`
`⇔ ((x-1)(x+1))/(x(x+1)) + (1x)/(x(x+1)) = (2x+1)/(x(x+1))`
`⇒ (x-1)(x+1) + x = 2x + 1`
`⇔ x^2 - 1 + x = 2x + 1`
`⇔ x^2 + x - 2x = 2x`
`⇔ x^2 - x - 2 = 0`
`⇔ (x^2+x) + (-2x-2) = 0`
`⇔ x(x+1) - 2(x+1) = 0`
`⇔ (x+1)(x-2) = 0`
`⇔`\(\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-1(KTM)\\x=2(TM)\end{array} \right.\)
Vậy `x \in {2}`
