Ta có: \(4x^2+12x+9\)

\(=4x^2+6x+6x+9\)

\(=2x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)^2\)

\(x^3+27x+\left(x+3\right)\left(x-9\right)\)
⇒\(x^3+27x+x^2-6x-27\)
⇒\(x^3+x^2+21x-27\)
Chịu

Sửa đề: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

\(=\left(x+3\right)\left(x^2-2x\right)=x\cdot\left(x-2\right)\left(x+3\right)\)

\(=x^4-x^3+3x^3-3x^2+3x^2-3x+9x-9\\ =\left(x-1\right)\left(x^3+3x^2+3x+9\right)\\ =\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)

\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+3x+9\right)\)

\(=\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)

\(=\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)

\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

3 . (x+3) - x²+9

= 3.(x+3)-(x²-9)

=3.(x+3)-[(x-3).(x+3)]

=(x+3).[3-(x-3)]

=(x+3).(3-x+3)

=(x+3).(9-x)

nhân tử gì ở đây

\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(2x-5\right)\\ =\left(x-3\right)\left(x+3+2x-5\right)\\ =\left(x-3\right)\left(3x-2\right)\)

\(x^2-9-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3-5+2x\right)=\left(x-3\right)\left(3x-2\right)\)

\(x^2-9-\left(x-3\right)\left(5-2x\right)\\ =\left(x^2-9\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3-5+2x\right)\\ =\left(x-3\right)\left(3x-2\right)\)

y^2(y+1)=9

nghiem vo ty roi xem lai de co nham ko?  lop 8 gi ra de kho vay 

(y+a)[y^2+(1-a)y-9/a]

a trong khoang (1,2) chin xac hon a la nghiem cua a^3+a^2-9=0 (lop 8 ko kho nhu vay, giai duoc nhung ban khong hieu duoc)

Bạn xem lại nhé. Đa thức đã cho không phân tích được thành nhân tử nữa. Chỉ thu gọn được thành $8x^3+50x^2-55x+10$

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)