Phân tích đa thức thành nhân tử
6x3y2-9x2y3+12x2y2
2x.(x-1)+3(1-x)
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10 tháng 10 2021
a) \(4\left(x+1\right)^3-x-1=4\left(x+1\right)^3-\left(x+1\right)=\left(x+1\right)\left[4\left(x+1\right)^2-1\right]=\left(x+1\right)\left[2\left(x+1\right)-1\right]\left[2\left(x+1\right)+1\right]=\left(x+1\right)\left(2x+1\right)\left(2x+3\right)\)
b) \(5x\left(x-3\right)+\left(3-x\right)^2-\left(x-3\right)=5x\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)=\left(x-3\right)\left(5x+x-3-1\right)=\left(x-3\right)\left(6x-4\right)=2\left(x-3\right)\left(3x-2\right)\)
c) \(9x^2y^3-3x^4y^2-6x^3y^2+16xy^4=xy^2\left(9xy-3x^3-6x^2+16y^2\right)\)
12 tháng 12 2023
a: \(9x^2y^3\left(3x-4y\right)+15x^3y^2\left(4y-3x\right)\)
\(=3x^2y^2\cdot\left(3x-4y\right)\cdot3y-3x^2y^2\cdot\left(3x-4y\right)\cdot5x\)
\(=3x^2y^2\left(3x-4y\right)\left(3y-5x\right)\)
b: \(4x^2+6x-9y^2-9y\)
\(=\left(4x^2-9y^2\right)+\left(6x-9y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
19 tháng 11 2021
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\\ =\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\\ =\left(x^2+3x+1\right)^2\)
NH
2
3 tháng 8 2015
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
HN
2
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 11 2021
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
24 tháng 11 2021
x(x+1)(x+2)(x+3)+1
= [x(x+3)][(x+1)(x+2)]+1
=(x2+3x)(x2+3x+2)+1
Đặt x2+3x+1=y, ta có:
(y-1)(y+1)+1
=y2-1+1
=y2
Thay y=x2+3x+1, lại có:
(x2+3x+1)2
H9
HT.Phong (9A5)
CTVHS
29 tháng 6 2023
\(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
a,\(6x^3y^2-9x^2y^3+1^2x^2y^2\)
\(=x^2y^2\left(6x-9y+1\right)\)
b,\(2x\left(x-1\right)+3\left(1-x\right)\)
\(=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)\)
a,
\(6x^3y^2-9x^2y^3+1\cdot x^2\cdot y^2\)
\(=x^2y^2\left(6x-9y+1\right)\)
b,
\(2x\left(x-1\right)+3\left(1-x\right)\)
\(=2x\left(x-1\right)+3\cdot-1\left(x-1\right)\)
\(=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)\)