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. Giúp nha
Và các bạn làm cho mình cả những bài mà mình đã gửi lên nữa nhé ! Thanks 
a) \(a^4+4\)
\(=a^4+4a^2+4-4a^2\)
\(=\left(a^2+2\right)^2+\left(2a\right)^2\)
\(=\left(a^2+2a+2\right)\left(a^2-2a+2\right)\)
b) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
(x80+x40+1)/(x20+x10+1)
=x60+x30
=x30(x2+1)
HIHI ÁNH BÉO TỚ LÀM BỪA 
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
= \(\left[\left(x+y+z\right)-\left(x+y\right)\right]^2\)
= \(z^2\)
Ta có:(x + y + z)2 - 2(x + y + z) (x + y) + (x + y)2
=[(x+y+z)-(x+y)]2=z2
, bạn nào biết thì giúp mình sửa nhé
a, Theo bài ra ta có:
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
b, theo bài ra ta có:
\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)
\(=x^2\left(x-3\right)-2x\left(x-3\right)-3\left(x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x-3\right)\)
c,Theo bài ra ta có:
\(=x^3+5x^2+3x^2+15x+2x+10\)
\(=x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2+3x+2\right)\)
\(=\left(x+5\right)\left(x^2+x+2x+2\right)=\left(x+5\right)\left(x\left(x+1\right)+2\left(x+1\right)\right)\)
\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)
CHÚC BẠN HỌC TỐT...........
a) \(x^3-3x+2\)
= \(x^3-x^2+x^2-x-2x+2\)
= \(x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x-2\right)\)
= \(\left(x-1\right)\left(x^2+2x-x-2\right)\)
= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(x-1\right)\)
= \(\left(x-1\right)^2\left(x+2\right)\)
b) \(x^3-5x^2+3x+9\)
= \(x^3+x^2-6x^2-6x+9x+9\)
= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-6x+9\right)\)
= \(\left(x+1\right)\left(x-3\right)^2\)
c) \(x^3+8x^2+17x+10\)
= \(x^3+x^2+7x^2+7x+10x+10\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10\right)\)
= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)
= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
d) \(x^3-3x^2+6x+4\)
Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:
\(x^3+3x^2+6x+4\)
= \(x^3+x^2+2x^2+2x+4x+4\)
= \(x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+2x+4\right)\)
Học tốt nhé :))
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
\(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)