phân tích đa thức thành nhân tử: (x^2-2x+1)^3+y^6
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\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Lời giải:
$2x(y-1)-3(1-y)=2x(y-1)+3(y-1)=(y-1)(2x+3)$
$3x^6-6=3(x^6-2)$
1: \(x\left(x-1\right)+\left(1+x\right)^2\)
\(=x^2-x+x^2+2x+1\)
\(=2x^2+x+1\)
Đa thức này ko phân tích được nha bạn
2: \(\left(x+1\right)^2-3\left(x+1\right)\)
\(=\left(x+1\right)\cdot\left(x+1\right)-\left(x+1\right)\cdot3\)
\(=\left(x+1\right)\left(x+1-3\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
3: \(2x\cdot\left(x-2\right)-\left(x-2\right)^2\)
\(=2x\left(x-2\right)-\left(x-2\right)\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
4: \(3x\left(x-1\right)^2-\left(1-x\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^2\cdot\left(x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(3x+x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(4x-1\right)\)
5: \(3x\left(x+2\right)-5\left(x+2\right)^2\)
\(=\left(x+2\right)\cdot3x-\left(x+2\right)\cdot\left(5x+10\right)\)
\(=\left(x+2\right)\left(3x-5x-10\right)\)
\(=\left(-2x-10\right)\left(x+2\right)\)
\(=-2\left(x+5\right)\left(x+2\right)\)
6: \(4x\left(x-y\right)+3\left(y-x\right)^2\)
\(=4x\left(x-y\right)+3\left(x-y\right)^2\)
\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)
\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)
\(=\left(x-y\right)\left(7x-3y\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)
ta có :(x^2-2x+1)^3+y^6=(x-1)^6+y^6 Còn lại tự hiểu
ta có :(x^2-2x+1)^3+y^6=(x-1)^6+y^6......Tự giải nốt