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

#)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:

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

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

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

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

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

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

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

cảm ơn

a: \(x^4-2x^3+x^2-2x\)

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

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

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

b: \(x^4+x^3-8x-8\)

\(=\left(x^4+x^3\right)-\left(8x+8\right)\)

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3-8\right)\)

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

\(b,x^3-2x^2-4xy^2+x\)

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

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

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

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

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

\(---\)

\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)

Đặt \(y=x^2+7x+10\), thay vào (1) ta được:

\(y\left(y+2\right)-8\)

\(=y^2+2y+1-9\)

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

\(=\left(y+1-3\right)\left(y+1+3\right)\)

\(=\left(y-2\right)\left(y+4\right)\)

\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)

\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

#Ayumu

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

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

phân tích ra  ii chj làm vậy ai chả lm đc

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

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

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

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

\(P=t^2-1\)

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

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

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

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

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