giải hộ mình với mọi người

Lời giải:

` 1) ` ` x^2 + 7x + 12 = 0 `

` <=> ` ` x^2 + 3x + 4x + 12 = 0 `

` <=> ` ` x(x + 3) + 4(x + 3) = 0 `

` <=> ` ` (x + 3)(x + 4) = 0 `

` <=> ` $\left[ \begin{array}{l}x+3=0\\x+4\end{array} \right.$

` <=> ` $\left[ \begin{array}{l}x=-3\\x=-4\end{array} \right.$

Vậy ` x = -3 ` hoặc ` x = -4 `

` 2) ` ` x^2 - 5x - 14 = 0 `

` <=> ` ` x^2 + 2x - 7x - 14 = 0 `

` <=> ` ` x(x + 2) - 7(x + 2) = 0 `

` <=> ` ` (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 = -2 ` hoặc ` x = 7 `

` 3) ` ` x^4 + 5x^2 = 6 `

` <=> ` ` x^4 - x^2 + 6x^2 - 6 = 0 `

` <=> ` ` x^2 (x^2 - 1) + 6(x^2 - 1) = 0 `

` <=> ` ` (x^2 + 6)(x^2 - 1) = 0 `

Vì ` x^2 + 6 >= 6 ` hay ` x^2 + 6 \ne 0 `

` => ` ` x^2 - 1 = 0 `

` <=> ` ` (x - 1)(x + 1) = 0 `

` <=> ` $\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.$

` <=> ` $\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.$

Vậy ` x = 1 ` hoặc ` x = -1 `

` 4) ` ` x^4 - 34x^2 + 225 = 0 `

` <=> ` ` x^4 - 9x^2 - 25x^2 + 225 = 0 `

` <=> ` ` x^2 (x^2 - 9) - 25(x^2 - 9) = 0 `

` <=> ` ` (x^2 - 9)(x^2 - 25) = 0 `

` <=> ` ` (x - 3)(x + 3)(x - 5)(x + 5) = 0 `

` <=> ` $\left[ \begin{array}{l}x-3=0\\x+3=0\\x-5=0\\x+5=0\end{array} \right.$

` <=> ` $\left[ \begin{array}{l}x=3\\x=-3\\x=5\\x=-5\end{array} \right.$

Vậy ` x \in {3;-3;5;-5} `