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Đáp án:
Giải thích các bước giải:
Bài 2:
$a)$
$5x(2x - 3) - 7(3 - 2x)$
$= 5x(2x - 3) + 7(2x - 3)$
$= (2x - 3)(5x + 7)$
$b)$
$x^4y - 3x^3y^2 + 3x^2y^3 - xy^4$
$= xy(x^3 - 3x^2y + 3xy^2 - y^3)$
$= xy(x - y)^3$
$c)$
$x^4 + 4y^4$
$= x^4 + 4x^2y^2 + 4y^2 - 4x^2y^2$
$= (x^2 + 2y^2) - (2xy)^2$
$= (x^2 + 2y^2 - 2xy)(x^2 + 2y^2 + 2xy)$
Bài 3:
$a)$
`3x(x - 1/2) = 0`
`<=>` \(\left[ \begin{array}{l}x=0\\x - \dfrac{1}{2} = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `S = {0; 1/2}.`
$b)$
$x^3 + x^2 - x - 1 = 0$
`<=> (x^3 + x^2) - (x + 1) = 0`
`<=> x^2(x + 1) - (x + 1) = 0`
`<=> (x + 1)(x^2 - 1) = 0`
`<=> (x + 1)^2(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 `S = {- 1; 1}.`
