phân tích đa thức thành nhân tử : (x^2-x+2).(x-1)-x^2.(1-x)^2-(2x+1).(1-x)^3
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HH
2
18 tháng 7 2021
\(\left(x-1\right)^2-2\left(x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left(x-1-2x-1\right)^2=\left(-x-2\right)^2=\left(x+2\right)^2\)
4 tháng 12 2023
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: \(4x\left(x-y\right)+3\left(y-x\right)^2\)
\(=4x\left(x-y\right)+3\left(x-y\right)^2\)
\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)
\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)
\(=\left(x-y\right)\left(7x-3y\right)\)
K
1
1 tháng 1 2022
\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
NH
3
NT
Nguyễn thành Đạt
CTVHS
24 tháng 9 2023
\(a)\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
Để đơn giản hơn cũng như là dễ nhìn hơn thì ta :
Đặt : \(x^2+2x=a\)
Do đó ta có đa thức :
\(a.\left(a+4\right)+3=a^2+4a+3\)
\(=a^2+a+3a+3\)
\(=a\left(a+1\right)+3\left(a+1\right)\)
\(=\left(a+1\right)\left(a+3\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(=\left(x+1\right)^2.\left(x^2+2x+3\right)\)
24 tháng 9 2023
Hoặc bạn có thể đặt \(x^2+2x+2=t\)
Thì \(P=\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)
\(P=\left(t-2\right)\left(t+2\right)+3\)
\(P=t^2-4+3\)
\(P=t^2-1\)
\(P=\left(t-1\right)\left(t+1\right)\)
\(P=\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
\(P=\left(x+1\right)^2\left(x^2+2x+3\right)\)
DV
0
DV
1
TL
1
14 tháng 7 2017
Ta có \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x\left(x^2-4\right)=1-4x^2-x^3+4x\)
\(=\left(1-x^4\right)+4x\left(1-x\right)=\left(1-x\right)\left(x^2+x+1\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
17 tháng 7 2023
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
\(\left(x^2-x+2\right)\left(x-1\right)-x^2\left(x-1\right)^2+\left(2x+1\right)\left(x-1\right)^3\)
\(=\left(x-1\right)\left[x^2-x+2-x^2\left(x-1\right)+\left(2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-x+2-x^3+x^2+2x^3-4x^2+2x+x^2-2x+1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+3\right)\)