Đáp án:
Bài 3:
a, `3x(x - 5) - 5 + x = 0`
` 3x(x - 5) - (5 - x) = 0`
` 3x(x - 5) + (x - 5) = 0`
`(3x + 1)(x - 5) = 0`
⇒\(\left[ \begin{array}{l}3x + 1 = 0\\x - 5 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}3x = -1\\x - 5 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = -\frac{1}{3}\\x = 5\end{array} \right.\)
Vậy `x \ in{ -1/3 ; 5}`
b, `x^2 - 4x - 3(4 - x) = 0`
`x(x - 4) + 3(x - 4) = 0`
`(3 + x)(x - 4) = 0`
⇒\(\left[ \begin{array}{l}3 + x = 0\\x - 4 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 0 - 3\\x = 0 + 4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = - 3\\x = 4\end{array} \right.\)
Vậy `x \ in{ -3 ; 4}`
Bài 4:
a, ` 4x(x + 1) = 8(x + 1)`
` 4x(x + 1) - 8(x + 1) = 0`
`(4x - 8)(x + 1) = 0`
⇒\(\left[ \begin{array}{l}4x - 8 = 0\\x + 1 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}4x = 8\\x = 0 - 1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 2\\x = - 1\end{array} \right.\)
Vậy `x \in {2 ; -1}`
b, `x^2 - 6x + 8 = 0`
`x^2 - 6x + 9 - 1 = 0`
`(x - 3)^2 = 1`
⇒\(\left[ \begin{array}{l}x - 3 = 1\\x - 3 = -1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 1 + 3\\x = -1 + 3\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 4\\x = 2\end{array} \right.\)
Vậy `x \in{ 4 ; 2}`
Bài 5:
a, `x^3 + x^2 + x + 1 = 0`
`x^2(x + 1) + (x + 1) = 0`
`(x^2 + 1)(x + 1) = 0`
Mà `x^2 + 1 > 0`
`⇒ x + 1 = 0`
` x = 0 - 1`
` x = -1`
b, `2(x - 4) - 4x + x^2 =0`
`2(x - 4) - x(4 - x) = 0`
`2(x - 4) + x(x - 4) = 0`
`(2 + x)(x - 4) = 0`
⇒\(\left[ \begin{array}{l}2 + x = 0\\x - 4 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 0 - 2\\x = 0 + 4\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = - 2\\x = 4\end{array} \right.\)
Vậy `x \in {-2 ; 4}`
