Đáp án:
Giải thích các bước giải:
`9x^2-1=(3x+1)(4x+1)`
`-> 12x^2 + 7x + 1 = 9x^2 - 1`
`-> 12x^2 + 7x + 2 = 9x^2`
`-> 3x^2 + 7x + 2 = 0`
`-> (3x^2+x)+(6x+2) = 0`
`-> x(3x+1) + 2(3x+1) = 0`
`-> (3x+1)(x+2) = 0`
`->` \(\left[ \begin{array}{l}3x+1=0\\x+2=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-\dfrac13\\x=-2\end{array} \right.\)
Vậy `x \in {-1/3,2}`
`(x+7)(3x-1) = 49 - x^2`
`-> 3x^2 + 20x - 7 = 49 - x^2`
`-> 4x^2 + 20x - 7 = 49`
`-> 4x^2 + 20x - 56 = 0`
`-> 4(x^2+5x-14) = 0`
`-> 4(x-2)(x+7) = 0`
`->`\(\left[ \begin{array}{l}x-2=0\\x+7=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=2\\x=-7\end{array} \right.\)
Vậy `x \in {2,-7}`
`(2x+1)^2=(x-1)^2`
`-> 4x^2 + 4x + 1 = x^2 - 4x + 4`
`-> 4x^2 + 4x - 3 = x^2 - 4x`
`-> 4x^2+8x-3=x^2`
`-> 3x^2+8x-3=0`
`-> (3x^2-x) + (9x-3) = 0`
`-> x(3x-1) + 3(3x-1) = 0`
`-> (x+3)(3x-1) = 0`
`->`\(\left[ \begin{array}{l}x+3=0\\3x-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-3\\x=\dfrac13\end{array} \right.\)
Vậy `x \in {-3,1/3}`
`x^3-5x^2+6x =0`
`-> x(x^2-2x-3x+6) = 0`
`-> x[x(x-2)-3(x-2)] = 0`
`-> x(x-2)(x-3) =0`
`->`\(\left[ \begin{array}{l}x=0\\x-2=0\\x-3=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=0\\x=2\\x=3\end{array} \right.\)
Vậy `x \in {0,2,3}`
