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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

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

\(0,125.\left(a+1\right)^3-1=\left[0,5\left(a+1\right)\right]^3-1^3=\left(0,5a+0,5-1\right)\left(0,5a+0,5+1\right)=\left(0,5a-0,5\right)\left(0,5a+1,5\right)=0,5\left(a-1\right).0,5\left(a+0,3\right)=0,25\left(a-1\right)\left(a+3\right)\)

\(0,125.\left(a+1\right)^3-1=\left(0,5a+0,5\right)^3-1=\left(0,5a+0,5-1\right)\left[\left(0,5a+0,5\right)^2+\left(0,5a+0,5\right)+1\right]=0,5\left(a-1\right)\left(0,25a^2+0,5a+0,25+0,5a+0,5+1\right)=\left(0,5a-0,5\right)\left(0,25a^2+a+1,75\right)\)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

`4a+1(a<=0=>-a>=0)`

`=1-4(-a)`

`=1-(2sqrt{-a})^2`

`=(1-2sqrt{-a})(1+2sqrt{-a})`

Với a nhỏ hơn 0 nhá 

\(4a+1=\left(2\sqrt{-a}-1\right)\left(2\sqrt{-a}+1\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

 ta có a^4 + a^2 + 1 = a ^ 4 + 2a^2 + 1 - a^2 =  (a^2+1)^2 - a^2 = (a^2- a + 1)(a^2 + a + 1)

\(a^4+a^2+1\)

\(=a^4+2a^2+1-a^2\)

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

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

\(a^{10}+a^5+1\)

\(=\left(a^{10}-a\right)+\left(a^5-a^2\right)+\left(a^2+a+1\right)\)

\(=a\left(a^3-1\right).\left(a^6+a^3+1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=a\left(a-1\right)\left(a^2+a+1\right)+\left(a^6+a^3+1\right)+a^2\left(a-1\right)\left(a^2+a+1\right)\)+  (a²+a+1) 

Đến đây rùi thì tự làm tiếp nha

\(a^{10}+a^5+1\)

\(=\left(a^{10}+a^9+a^8\right)-\left(a^9+a^8+a^7\right)+\left(a^7+a^6+a^5\right)-\left(a^6+a^5+a^4\right)+\left(a^5+a^4+a^3\right)-\left(a^3+a^2+a\right)+\left(a^2+a+1\right)\)\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)