Đáp án + Giải thích các bước giải:
`1` , `(2x - 1)^3 = (2x - 1)^5`
`(2x - 1)^5 - (2x - 1)^3 = 0`
`(2x - 1)^3 . [ (2x - 1)^2 - 1] = 0`
⇒\(\left[ \begin{array}{l}(2x - 1)^3 = 0\\(2x - 1)^2- 1= 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x - 1 = 0\\2x - 1 = 1\\ 2x - 1 = -1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x = 1\\2x =2\\ 2x =0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = \frac{1}{2}\\x=1\\ x =0\end{array} \right.\)
Vậy `x \in {1/2 ; 1 ; 0}`
`2` , `x^2 + 8x - 9 = 0`
` x^2 + 9x - x - 9 = 0`
` x(x + 9) - (x + 9) = 0`
`(x - 1)(x + 9) = 0`
⇒\(\left[ \begin{array}{l}x-1=0\\x+9=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1\\x=-9\end{array} \right.\)
Vậy `x \in {1 ; -9}`
`3` , `x^2 + 11x - 12 = 0`
` x^2 + 12x - x - 12 = 0`
` x(x + 12) - (x + 12) = 0`
` (x - 1)(x + 12) = 0`
⇒\(\left[ \begin{array}{l}x-1=0\\x+12=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1\\x=-12\end{array} \right.\)
Vậy `x \in {1 ; -12}`
