![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(e,x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y-1\right)^2\)
Bài 2:
\(b,2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.4x^3-8x^2+4xy^3=4x\left(x^2-8x+y^3\right)\)
\(b.x^2+2xy+y^2-36=\left(x+y\right)^2-36=\left(x+y-6\right)\left(x+y+6\right)\) \(c.x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\) \(d.x^2-5x+2xy-5y+y^2=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\) \(e.49+2xy-x^2-y^2=-\left(x^2-2xy+y^2-49\right)=-\left[\left(x-y\right)^2-49\right]=-\left(x-y-7\right)\left(x-y+7\right)\) \(f.3x^2-6x+3-3y^2=3\left(x^2-2x-y^2+1\right)\)
\(g.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)\left(x+1\right)\)
\(h,\) giống câu f.
\(i.x^3-2x^2y+xy^2-64x=x\left(x^2-2xy+y^2-64\right)=x\left[\left(x-y\right)^2-64\right]=x\left(x-y-8\right)\left(x-y+8\right)\) \(k.3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ A = 2x2 + y2 - 2xy - 2x + 3
= (x2 - 2xy + y2) + (x2 - 2x + 1) + 2
= (x - y)2 + (x - 1)2 + 2\(\ge2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2 nek :
x2(2 − x) − x + 2 = 0
x2(2 - x) + (2 - x) = 0
(x2 +1)(2 - x) = 0
\(\Rightarrow\) \(\left[{}\begin{matrix}2-x=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\) \(\left[{}\begin{matrix}x=2\\x=\sqrt{-1}\end{matrix}\right.\)
Chúc bn học tốt
a) \(2x^2-6x=2x\left(x-3\right)\)
b) \(x^2-y^2-3x+3y=\left(x^2-y^2\right)-\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
c) \(x^2+2xy+y^2-9z^2=\left(x+y\right)^2-\left(3z\right)^2\)
\(=\left(x+y-3z\right)\left(x+y+3z\right)\)
d) \(2x^2-5x-3=2x^2-2x+3x-3\)
\(=\left(2x^2-2x\right)+\left(3x-3\right)\)
\(=2x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(2x+3\right)\left(x-1\right)\)
e) \(2x^2+2xy=2x\left(x+y\right)\)
g) \(x^2-x-6=x^2-3x+2x-6\)
= \(\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x+2\right)\left(x-3\right)\)
Bài 2: \(x^2\left(2-x\right)-x+2=0\)
\(x^2\left(2-x\right)+\left(2-x\right)=0\)
\(\left(x^2+1\right)\left(2-x\right)=0\)
=> \(x^2+1=0hay2-x=0\)
=> x không tồn tại hay x = 2
=> x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích đa thức thành nhân tử
a. \(12x^3y-24x^2y^2+12xy^3\)
\(=12xy\left(x^2-2xy+y^2\right)\)
\(=12xy\left(x-y\right)^2\)
b. \(x^2-6x+xy-6y\)
\(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x+y\right)\left(x-6\right)\)
c. \(2x^2+2xy-x-y\)
\(=x\left(2x-1\right)+y\left(2x-1\right)\)
\(=\left(x+y\right)\left(2x-1\right)\)
d. \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
e. \(x^2-2xy-x^2-4y^2\)
\(=-2xy-4y^2\)
\(=-2y\left(x+2y\right)\)
g. \(x^2-2x+1-16\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-1-4\right)\left(x-1+4\right)\)
\(a,\left(x^2-1\right)\cdot\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
\(b,\left(x+3y\right)\cdot\left(x^2-2xy+y\right)=x^3-2x^2y+xy+3x^2y-6xy^2+3y^2\)
\(=x^3+x^2y+xy+3y^2\)
a) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^2\left(x^2-1\right)+2x\left(x^2-1\right)\)
\(=x^4-x^2+2x^3-2x\)
\(=x^4+2x^3-x^2-2x\)
b) \(\left(x+3y\right)\left(x^2-2xy+y\right)\)
\(=x\left(x^2-2xy+y\right)+3y\left(x^2-2xy+y\right)\)
\(=x^3-2x^2y+xy+2x^2y-6xy^2+3y^2\)
\(=x^3-6xy^2+3y^2+x\)