Đáp án:

 `1, x^3+y^3+z^3-3xyz`

`=(x+y)^3-3xy(x+y)+z^3-3xyz`

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

`=(x+y+z)(x^2+y^2+z^2+2zy-xz-yz-3xy)`

`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`

`2, 3(x-3)(x+7)+(x-4)^2+48`

`=3(x^2-3x+7x-21)+x^2-8x+16+48`

`=3x^2+12x-63+x^2-8x+64`

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

`=(2x+1)^2`

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

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

`<=> (x+1)(x+1-1)=0`

`<=> x(x+1)=0`

`<=>`\(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\) 

`4, x^3-0,25=0`

`<=> x^3=1/4`

`<=> x=∛(1/4)` (mik nghĩ câu này sai đề ròi)

1.x³+y³+z³-3xyz=(x+y)³-3xy(x+y)+z³-3xyz(DỰA VÀO BÀI 31 SGK)

                         =(x+y)³+z³-[3xy(x+y)-3xyz]

                         =[(x+y)³+z³]-3xy(x+y+z)

                         =(x+y+z)[(x+y)²-(x+y)z+z²]-3xy(x+y+z)

                         =(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)

                         =(x+y+z)(x²+y²+z²-xy-xz-yz)

2.x+1=(x+1)²⇒(x+1)²-(x+1)=(x+1)(x+1)-(x+1)1

                                            =(x+1)(x+1-1)=(x+1)x