1 . Phân tích đa thức thành nhân tử :
a ) x^3+2x^2+x-16xy^2
b ) x^2+2+xy-9+y^2
c ) x^2-3xy-10y^2
2 . Tìm x biết :
a ) x(x-2)-x+2=0
b ) x^2(x^2+1)-x^2-1 = 0
c ) 5x(x-3)^2-5(x-1)^3+15(x+2)(x-2) = 5
1 . Phân tích đa thức thành nhân tử :
a ) x^3+2x^2+x-16xy^2
b ) x^2+2+xy-9+y^2
c ) x^2-3xy-10y^2
2 . Tìm x biết :
a ) x(x-2)-x+2=0
b ) x^2(x^2+1)-x^2-1 = 0
c ) 5x(x-3)^2-5(x-1)^3+15(x+2)(x-2) = 5
$Bài_{}$ $1:_{}$
$a)_{}$ $x^3+2x^2+x-16xy^2_{}$
$⇔x.(x^2+2x+1-16y^2)_{}$
$⇔x. [ (x+1)^2-16y^2] _{}$
$⇔x.(x+1-4y).(x+1+4y)_{}$
$b)_{}$ $x^2+2xy-9+y^2_{}$
$⇔(x^2+2xy+y^2)-9_{}$
$⇔(x+y)^2-3^2_{}$
$⇔(x+y-3)(x+y+3)_{}$
$c)_{}$ $x^2-3xy-10y^2_{}$
$⇔x^2+2xy-5xy-10y^2_{}$
$⇔x.(x+2y)-5y.(x+2y)_{}$
$⇔(x-5y)(x+2y)_{}$
$Bài_{}$ $3:_{}$
$⇔x.(x-2)-x+2=0_{}$
$⇔x.(x-2)-(x-2)=0_{}$
$⇔(x-1)(x-2)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
$b)_{}$ $x^2.(x^2+1)-x^2-1=0_{}$
$⇔x^2.(x^2+1)-(x^2+1)=0_{}$
$⇔(x^2-1).(x^2+1)=0_{}$
$⇔(x-1)(x+1)(x^2+1)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
$(Vì_{}$ $x^2+1=0_{}$ $là_{}$ $vô_{}$ $nghiệm)_{}$
$c)_{}$ $5x.(x-3)^2-5.(x-1)^3+15.(x+2)(x-2)=5_{}$
$⇔5x.(x^2-6x+9)-5.(x^3-3x^2+3x-1)+15.(x^2-4)=5_{}$
$⇔5x^3-30x^2+45x-5x^3+15×62-15x+5+15x^2-60=5_{}$
$⇔30x-55=5_{}$
$⇔30x=60_{}$
$⇔x=2_{}$