`1)`
`a)x^3-y^3=(x-y)(x^2+xy+y^2)`
`b)(1+x)(1-2x)-(x+1)(x-2)`
`=(x+1)(1-2x-x+2)`
`=(x+1)(3-3x)`
`c)(2x+1)^2+(x-3)^2`
`=(2x+1-x+3)(2x+1+x-3)`
`=(x+4)(3x-2)`
`d)x^2-4xy-5y^2`
`=x^2-5xy+xy-5y^2`
`=x(x-5y)-y(x-5y)`
`=(x-y)(x-5y)`
`e)x^6-y^6`
`=(x^3)^2-(y^3)^2`
`=(x^3-y^3)(x^3+y^3)`
`=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)`
`2)`
`a)x^2-2x=0`
`⇒x(x-2)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x-2=0=>x=2\end{array} \right.\)
Vậy `x∈{0;2}`
`b)x^3-x^2-2x=0`
`⇒x(x^2-x-2)=0`
`⇒x(x^2+2x-x-2)=0`
`⇒x[(x^2+2x)-(x+2)=0`
`⇒x[x(x+2)-(x+2)]=0`
`⇒x(x-1)(x+2)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x-1=0\\x+2=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=1\\x=-2\end{array} \right.\)
Vậy `x∈{0;1;-2}`
`3)`
`a)` Có `(x+y)^2=x^2+y^2+2xy`
`⇒x^2+y^2=(x+y)^2-2xy`
mà `x+y=2,x.y=1⇒2^2-2.1=2`
Vậy `x^2+y^2=2`
`b)` Có `x^3+y^3=x^3+y^3+3x^2y+3xy^2`
`(x+y)^3`
`=x^3+y^3+3x^2y+3xy^2`
`=x^3+y^3+3xy(xy)`
`⇒x^3+y^3=(x+y)^3-3xy(xy)`
mà `x+y=1,x.y=-3⇒x^3+y^3=1^3-3.(-3).1=10`
Vậy `x^3+y^3=10`
`c)(x-y)^2=x^2+y^2-2xy`
`⇒x^2+y^2=(x-y)^2-2xy`
mà `xy=-1, x-y=2` `⇒(x-y)^2-2xy=4-2(-1)=6`
`(x^2+y^2)^2=x^4+y^4+2x^2y^2`
`⇒x^4+y^4=(x^2y^2)^2-2x^2y^2=6^2-2(-1)^2=36-2=34`
Vậy `x^4+y^4=34`
