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Bài 1:
a.
`x^2y - y`
`= y(x^2 - 1)`
`= y(x - 1)(x + 1)`
b.
`x^2 - 25y^2 + 2x + 1`
`=(x^2 + 2x + 1) - (5y)^2`
`=(x + 1)^2 - (5y)^2`
`=(x + 1 - 5y)(x + 1 + 5y)`
c.
`x^3 + 3x^2 - x - 3`
`=(x^3 - x) + (3x^2 - 3)`
`=x(x^2 - 1) + 3(x^2 - 1)`
`=(x^2 - 1)(x + 3)`
`=(x - 1)(x + 1)(x + 3)`
Bài 2:
a.
`x^2 - 5x = 0`
`⇔x(x - 5) = 0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\)
Vậy `x = {0; 5}`
b.
`3(x + 2) - x^2 + 4 = 0`
`⇔3(x + 2) - (x^2 - 4) = 0`
`⇔3(x + 2) - (x - 2)(x + 2) = 0`
`⇔(x + 2)(3 - x - 2) = 0`
`⇔(x + 2)(1 - x) = 0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\1-x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
Vậy `x = {-2; 1}`
c.
`x^2 - 5x + 6 = 0`
`⇔x^2 - 3x - 2x + 6 = 0`
`⇔(x^2 - 3x) - (2x - 6) = 0`
`⇔x(x - 3) - 2(x - 3) = 0`
`⇔(x - 3)(x - 2) = 0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
Vậy `x = {2; 3}`
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Bài 1/
a/ $x^2y-y$
$=y(x^2-1)=y(x+1)(x-1)$
b/ $x^2-25y^2+2x+1$
$=(x+1)^2-25y^2$
$=(x+1-5y)(x+1+5y)$
c/ $x^3+3x^2-x-3$
$=x^2(x+3)-(x+3)$
$=(x+3)(x^2-1)=(x+3)(x-1)(x+1)$
Bài 2/
a/ $x^2-5x=0$
$\leftrightarrow x(x-5)=0$
\(\leftrightarrow \left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\)
b/ $3(x+2)-x^2+4=0$
$\leftrightarrow 3(x+2)-(x^2-4)=0$
$\leftrightarrow 3(x+2)-(x-2)(x+2)=0$
$\leftrightarrow (x+2)(3-x+2)=0$
$\leftrightarrow (x+2)(1-x)=0$
\(\leftrightarrow\left[ \begin{array}{l}x+2=0\\1-x=0\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
c/ $x^2-5x+6=0$
$\leftrightarrow x^2-2x-3x+6=0$
$\leftrightarrow (x^2-2x)-(3x-6)=0$
$\leftrightarrow x(x-2)-3(x-2)=0$
$\leftrightarrow (x-3)(x-2)=0$
\(\leftrightarrow\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
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