Đáp án + giải thích các bước giải:
Bài `1`:
`a, 5x(x^3 - 3x + 4)`
` = 5x . x^3 - 5x . 3x + 5x . 4`
` = 5x^4 - 15x^2 + 20x`
`b, (x - 2y)(x + 2y)`
` = x^2 - (2y)^2`
` = x^2 - 4y^2`
`c, 3x(2x - 5) - (4x - 3)(2 - x)`
` = 6x^2 - 15x - (8x - 4x^2 - 6 + 3x)`
` = 6x^2 - 15x - 11x + 4x^2 + 6`
` = 10x^2 - 26x + 6`
Bài `2`:
`a, 4x^2 - 2x - 9y^2 - 3y`
` = (4x^2 - 9y^2) - (2x + 3y)`
` = [ (2x)^2 - (3y)^2] - (2x + 3y)`
` = (2x - 3y)(2x + 3y) - (2x + 3y)`
` = (2x + 3y)(2x - 3y - 1)`
`b, 3xy^2 - 3x^3 + 6x^2 - 3x`
` = 3x(y^2 - x^2 + 2x - 1)`
` = 3x[ y^2 - (x^2 - 2x + 1)]`
` = 3x[ y^2 - (x - 1)^2]`
` = 3x(y - x + 1)(y + x - 1)`
Bài `3`:
`a, x(x - 2) + 5(x - 2) = 0`
`<=> (x + 5)(x - 2) = 0`
`<=>` \(\left[ \begin{array}{l}x + 5 = 0\\x - 2 = 0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x = -5\\x = 2\end{array} \right.\)
Vậy `x \in {-5 ; 2}`
`b, x + 2 \sqrt{2} x^2 + 2x^3 = 0`
`<=> x(1 + 2 \sqrt{2} x + 2x^2) = 0`
`<=> x( \sqrt{2}x + 1)^2 = 0`
`<=>` \(\left[ \begin{array}{l}x =0\\(\sqrt{2}x + 1)^2 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x =0\\\sqrt{2}x + 1 = 0\end{array} \right.\)`
`<=>` \(\left[ \begin{array}{l}x =0\\x = - \dfrac{1}{\sqrt{2}}\end{array} \right.\)
Vậy `x \in {0 ; -1/(\sqrt{2} ) }`
