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

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

x³ + 2x² + 2x + 1 

= (x³ + 1) + (2x² + 2x)

= (x + 1)(x² + x + 1) + 2x(x + 1)

= (x + 1)(x² + x + 1 + 2x)

= (x + 1)(x² + 3x + 1)

 Chúc bạn học tốt

Ta có: \(x^3+2x^2+2x+1\)

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

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

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

#)Giải :

\(x^3-2x-4\)

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

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

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

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

cảm ơn nha!

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

  (2\(x\) - 1)² - (3\(x\))²

= (2\(x\) - 1 - 3\(x\)).( 2\(x\) - 1+ 3\(x\))

= (- \(x\) - 1).(5\(x\) - 1)

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

\(a)\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)

Để đơn giản hơn cũng như là dễ nhìn hơn thì ta :

Đặt : \(x^2+2x=a\)

Do đó ta có đa thức :

\(a.\left(a+4\right)+3=a^2+4a+3\)

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

\(=a\left(a+1\right)+3\left(a+1\right)\)

\(=\left(a+1\right)\left(a+3\right)\)

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

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

Hoặc bạn có thể đặt \(x^2+2x+2=t\)

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

\(P=\left(t-2\right)\left(t+2\right)+3\)

\(P=t^2-4+3\)

\(P=t^2-1\)

\(P=\left(t-1\right)\left(t+1\right)\)

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

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

a) \(\left(x^2+2x\right).\left(x^2+2x+4\right)+3\)

\(=x^4+4x^3+4x^2+4x^3+16x^2+16x\)

\(=x^4+8x^3+20x^2+16x\)

\(=\left(x^4+8x^3+20x^2+16x\right)+3\)

\(=x^4+8x^3+20x^2+16x+3\)

1) \(x\left(4x+1\right)\)

2) \(3\left(x-3y\right)\)

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

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

\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)