Đáp án: $a)$ \(\left[ \begin{array}{l}x=8/3\\x=2\end{array} \right.\)
$b)$ \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
$c)$ $a = 1; b = -1; c= 0$
Giải thích các bước giải:
$Bài$ $1: a)$ $(x - 3)^2 = 4x^2 - 20x + 25$
$<=> x^2 - 6x + 9 = 4x^2 - 20x + 25$
$<=> (x^2 - 4x^2) - (6x - 20x) + (9 - 25) = 0$
$<=> -3x^2 + 14x - 16 = 0$
$<=> -3x^2 + 8x + 6x - 16 = 0$
$<=> (6x -3x^2) + (8x - 16) = 0$
$<=> 3x.(2 -x) - 8.(2 - x) = 0$
$<=> (3x-8)(2 -x) = 0$
$<=>$ \(\left[ \begin{array}{l}3x-8=0\\2-x=0\end{array} \right.\)
$<=>$ \(\left[ \begin{array}{l}x=8/3\\x=2\end{array} \right.\)
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$b)$ $(x - 1)^2 - (2x + 1)^2 = 0$
$<=> x^2 - 2x + 1 - (4x^2 + 4x + 1) = 0$
$<=> x^2 - 2x + 1 - 4x^2 - 4x - 1 = 0$
$<=> (x^2 - 4x^2) + (- 2x - 4x) + (1 - 1) = 0$
$<=> -3x^2 - 6x = 0$
$<=> -3x.(x + 2) = 0$
$<=>$ \(\left[ \begin{array}{l}-3x=0\\x+2=0\end{array} \right.\)
$<=>$ \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
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$c)$ $(ay^2 + by + c)(y + 3) = y^3 + 2y^2 - 3y$
$<=> (ay^2 + by + c)(y + 3) = y^3 + 3y^2 - y^2 - 3y$
$<=> (ay^2 + by + c)(y + 3) = y^2.(y + 3) - y.(y + 3)$
$<=> (ay^2 + by + c)(y + 3) = (y^2-y)(y + 3)$
$<=> \dfrac{(ay^2 + by + c)(y + 3)}{(y + 3)}= \dfrac{(y^2-y)(y + 3)}{(y + 3)}$
$=> ay^2 + by + c = y^2 - y$
$=> a = 1; b = -1; c= 0$
