Bài 7-bài8 $i)3x+3y-x^2-2xy-y^2$ $j)x^4y^4+64$,$k)x^3+3x^2y+3xy^2+y^3-x-y$ $1)x^8+x+1$,$m)x^3+y^3-2(x^2-y^2)$ $n)(x^2+y)^2-2x^2-2y+1$,$o)x^2+5x+6$ $p)(x^2+x+1)(x^2+x+2)-6$,Bài 7. Phân tích đa thức thành nhân tử:,$a)x^3y+x-y-1$ $b)x^2(x-2)+4(2-x)$,$c)x^3-x^2-20x$ $d)(x^2+1)^2-(x+1)^2$,$e)6x^2-7x+2$ $f)x^4+8x^2+12$,$g)(x^3+x+1)(x^3+x)-2$ $h)(x+1)(x+2)(x+3)(x+4)+1$,$i)-(x^2+2)^2+4x(x^2+2)-3x^2$ $j)81x^4+4y^4$,Bài 8. Tìm x, biết:,$a)(x+2)^2-x(x+3)=2$,$b)(x+2)(x-2)-(x+1)^2=7$,$c)6x^2-(2x+1)(3x-2)=1$,$d)(x+2)(x+3)-(x-2)(x+1)=2$,$e)6(x-1)(x+1)-(2x-1)(3x+2)+3=0$,$ extcircled1x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$,$g)(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$,$1)(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$,$(x+1)-6(x-1)^2=23$

Question

Bài 7-bài8 $i)3x+3y-x^2-2xy-y^2$ $j)x^4y^4+64$,$k)x^3+3x^2y+3xy^2+y^3-x-y$ $1)x^8+x+1$,$m)x^3+y^3-2(x^2-y^2)$ $n)(x^2+y)^2-2x^2-2y+1$,$o)x^2+5x+6$ $p)(x^2+x+1)(x^2+x+2)-6$,Bài 7.   Phân tích đa thức thành nhân tử:,$a)x^3y+x-y-1$ $b)x^2(x-2)+4(2-x)$,$c)x^3-x^2-20x$ $d)(x^2+1)^2-(x+1)^2$,$e)6x^2-7x+2$ $f)x^4+8x^2+12$,$g)(x^3+x+1)(x^3+x)-2$ $h)(x+1)(x+2)(x+3)(x+4)+1$,$i)-(x^2+2)^2+4x(x^2+2)-3x^2$ $j)81x^4+4y^4$,Bài 8.   Tìm x, biết:,$a)(x+2)^2-x(x+3)=2$,$b)(x+2)(x-2)-(x+1)^2=7$,$c)6x^2-(2x+1)(3x-2)=1$,$d)(x+2)(x+3)-(x-2)(x+1)=2$,$e)6(x-1)(x+1)-(2x-1)(3x+2)+3=0$,$	extcircled1x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$,$g)(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$,$1)(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$,$(x+1)-6(x-1)^2=23$

Accepted Answer

Bài 7:
a/
$\displaystyle \begin{array}{{>{\displaystyle}l}}
x^{3} y+x-y-1\
=\left( x^{3} -1 ight) .y+( x-1)\
=y( x-1)\left( x^{2} +x+1 ight) +( x-1)\
=( x-1)\left( x^{2} y+xy+y+1 ight)
\end{array}$
b/
$\displaystyle \begin{array}{{>{\displaystyle}l}}
x^{2}( x-2) +4( 2-x)\
=x^{2}( x-2) -4( x-2)\
=\left( x^{2} -4 ight)( x-2)\
=( x-2)^{2}( x+2)
\end{array}$
c/
$\displaystyle \begin{array}{{>{\displaystyle}l}}
x^{3} -x^{2} -20x\
=x\left( x^{2} -x-20 ight)\
=x( x-5)( x+4)
\end{array}$
d/
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\left( x^{2} +1 ight)^{2} -( x+1)^{2}\
=\left( x^{2} +1-x-1 ight)\left( x^{2} +1+x+1 ight)\
=\left( x^{2} -x ight)\left( x^{2} +x+2 ight)\
=x( x-1)\left( x^{2} +x+2 ight)
\end{array}$