Bài 1:

`a)(x+y)(x^2-xy+y^2)-(-x+y)(x^2+xy+y^2)`

`=x^3+y^3-(y^3-x^3)`

`=x^3+y^3-y^3+x^3`

`=2x^3`

`b)(x+y)^3-(x-2y)^3`

`=(x+y-x+2y)[(x+y)^2+(x+y)(x-2y)+(x-2y)^2]`

`=3y (x^2+2yx+y^2+x^2+xy-2xy-2y^2+x^2-4xy+4y^2)`

`=3y (3x^2-3xy+3y^2)`

`c)(3x+4y-5z)(3x-4y+5z)`

`=[3x+(4y-5z)][3x-(4y-5z)]`

`=(3x)^2-(4y-5z)^2`

`=9x^2-16xy^2+40yz-25z^2`

`d)(x^2+4x+3)^2+(4x+2)^2-2(x^2+3x+1)(4x+2)`

`=(x^2+4x+3-4x-2)^2`

`=(x^2+1)^2`

`=x^4+2x^2+1`

Bài 2:

`a)(x+2)^2+(x+3)^2-2(x+2)(x+3)=16`

`<=>(x+3-x-2)^2=16`

`<=>1^2=16`(vô lý)

Vậy phương trình vô no 

`b)(x+2)(X^2-2x+4)-x(x^2-5)=15`

`<=>x^3+8-x^3+5x=15`

`<=>5x=15-8`

`<=>5x=7`

`<=>x=7/5`

Vậy `x=7/5`

`c)(x-1)^3+(2-x)(4-2x+x^2)+3x(x+2)=17`

`<=>x^3-3x^2+3x-1+8-x^3+3x^2+6x=17`

`<=>9x+7=17`

`<=>9x=10`

`<=>x=10/9`

Vậy `x=10/9`

`d)x^2-2x+5+y^2-4y=0`

`<=>(x^2-2x+1)+(y^2-4y+4)=0`

`<=>(x-1)^2+(y-2)^2=0`

Ta có `(x-1)^2>=0;(y-2)^2>=0`

`=>(x-1)^2+(y-2)^2>=0`

Dấu `'='` xảy ra `<=>{(x-1=0),(y-2=0):}`

`<=>{(x=1),(y=2):}`

`