\(x^3+3x^2+3x+2\)

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

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

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

\(x^3+3x^2 +3x+2\)

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

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

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

\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)

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

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

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

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

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

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

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

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

mình biết nội quy rồi nên đưng đăng nội quy

ai chơi bang bang 2 kết bạn với mình

mình có nick có 54k vàng đang góp mua pika 

ai kết bạn mình cho

\(A=\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

Đặt \(t=x^2+3x+1\) thì A thành

\(t\left(t-4\right)-5=t^2-4t-5\)

\(t^2-5t+t-5=t\left(t-5\right)+\left(t-5\right)\)

\(=\left(t-5\right)\left(t+1\right)=\left(x^2+3x+1-5\right)\left(x^2+3x+1+1\right)\)

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

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

đặt a=x^2+3x+1

phương trình đã cho thành phương trình: a(a-4)-5

=a^2-4a-5

=a^2+a-5a-5

= a(a+1)-5(a+1)

=(a-5)(a+1)

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

=(x-1)(x+1)(x+2)(x+4)

mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)

                    \(=[(2x)^2-3^2](4x^2+9)\)

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

b, \(x^3-3x^2+3x-1=(x-1)^3\)

\(x^2-2x+1=(x-1)^2\)

c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)

\(=(6x^2+3x+1)(3x^3+1)\)

câu c bạn đánh sai 1 dấu phép toán kìa!!!!

a; \(\left(16x^4-81\right):\left(2x+3\right)\)

\(=\frac{\left(4x^2-9\right)\left(4x^2+9\right)}{2x+3}\)

\(=\frac{\left(2x-3\right)\left(2x+3\right)\left(4x_{}^2+9\right)}{2x+3}=\left(2x-3\right)\left(4x^2+9\right)\)

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

\(=\frac{\left(x-1\right)^3}{\left(x-1\right)^2}=x-1\)

c: Sửa đề: \(\left(18x^5+9x^4+3x^3+6x^2+3x+1\right):\left(6x^2+3x+1\right)\)

\(=\frac{3x^3\left(6x^2+3x+1\right)+\left(6x^2+3x+1\right)}{6x^2+3x+1}\)

\(=\frac{\left(6x^2+3x+1\right)\left(3x^2+1\right)}{6x^2+3x+1}=3x^2+1\)

x² - 3x + 2 = x( x - 3 ) + 2

j` nữa thì bó tay @@  

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

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

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