`5/2 - 1/2 + 2/3`

`=4/2 + 2/3`

`=2+ 2/3`

`=6/3 + 2/3`

`=8/3`

`1/4 + 2/3 xx 1/5`

`=1/4 + 2/15`

`=15/60 + 8/60`

`=23/60`

`b,(a+b)^3 - (a-b)^3 - 6a^2 b`

`=(a^3+3a^2 b+3ab^2 +b^3)-(a^3-3a^2 b+3ab^2 -b^3)-6a^2 b`

`=a^3+3a^2 b+3ab^2 +b^3-a^3 +3a^2 b-3ab^2 +b^3 -6a^2 b`

`=2b^3`

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

`=[(x^2)^3 -3(x^2)^2 .1 + 3x^2 .1^2 -1^3)-[(x^2)^3 -1^3]`

`=(x^6 -3x^4 + 3x^2 -1)-(x^6 -1)`

`=x^6 -3x^4 + 3x^2 -1-x^6 +1`

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

`5/2 - 1/2 + 2/3`

`=4/2 + 2/3`

`=2+ 2/3`

`=6/3 + 2/3`

`=8/3`

`1/4 + 2/3 xx 1/5`

`=1/4 + 2/15`

`=15/60 + 8/60`

`=23/60`

dùng dấu = ko phải tương đương

ghi tham khảo vào

\(1,x^2-y^2+4x-4y\\ =\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\\ =\left(x-y\right)\left(x+y+4\right)\\ 2,x^2-6x-7\\ =x^2-7x+x-7\\ =x\left(x-7\right)+\left(x-7\right)\\ =\left(x-7\right)\left(x+1\right)\\ 3,2x^3-8x^2+8x\\ =2x\left(x^2-4x+4\right)\\ =2x\left(x-2\right)^2\\ 4,2x^2+3x-5\\ =2x^2+5x-2x-5\\ =x\left(2x+5\right)-\left(2x+5\right)\\ =\left(x-1\right)\left(2x+5\right)\)

`5,` xem lại đề 

\(6,6x^2y-3xy^2\\ =3xy\left(2x-y\right)\\ 7,x^2-y^2+10x+25\\ =\left(x^2+10x+25\right)-y^2\\ =\left(x+5\right)^2-y^2\\ =\left(x-y+5\right)\left(x+y+5\right)\)

ít tham khảo giùm nếu ko bt lm thì thôi đừng lm

Để đa thức `x^3 + x^2 +a -x vdots (x+1)^2` thì:

`x^3 +x^2 +a -x=(x+1)^2 .f(x)`(*)

Thay `x=-1` vào (*) ta có:
`(-1)^3 + (-1)^2 + a - (-1) = (-1+1)^2 . f(x)`

`<=> -1 +1 + a +1 = 0^2 . f(x)`

`<=>  a +1 = 0`

`<=> a=-1`

 Vậy `a=-1`