\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)

Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)

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

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

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

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

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

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

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

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

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

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

 Xét đa thức \(g\left(x\right)=x^5+x^4+x^3+2x^2+x+1\) có bậc 5 là số lẻ. Khi đó giả sử tồn tại 2 đa thức \(h\left(x\right)\) và \(j\left(x\right)\) hệ số nguyên sao cho:

 \(g\left(x\right)=h\left(x\right).j\left(x\right)\). Khi đó 1 trong 2 đa thức \(h\left(x\right),j\left(x\right)\) phải có bậc lẻ (vì nếu cả 2 đều bậc chẵn thì thành thử bậc của \(g\left(x\right)\) phải chẵn, mâu thuẫn theo trên).

 Không mất tổng quát, giả sử đa thức \(h\left(x\right)\) có bậc lẻ. Khi đó nếu nó có nghiệm hữu tỉ thì gọi nghiệm hữu tỉ này là \(x=\dfrac{p}{q}\left(p,q\inℤ;\left(p,q\right)=1\right)\) thì \(p|1,q|1\) nên \(x=\pm1\). Thử lại, ta thấy 2 nghiệm này đều không thỏa mãn.

 Do đó, \(g\left(x\right)\) không có nghiệm vô tỉ nên ta không thể phân tích tiếp \(f\left(x\right)\) thành nhân tử được nữa.

h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)

i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)

a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)

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

Lời giải:
a.

$x^4+10x^3+26x^2+10x+1$

$=(x^4+10x^3+25x^2)+x^2+10x+1$

$=(x^2+5x)^2+2(x^2+5x)+1-x^2$

$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$

$=(x^2+4x+1)(x^2+6x+1)$

b.

$x^4+x^3-4x^2+x+1$

$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$

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

$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$

$=(x-1)(x^3+2x^2-2x-1)$

$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$

$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$

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

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

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

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