Bài 2:
a: \(4x^2+3x=0\)
=>x(4x+3)=0
=>\(\left[\begin{array}{l}x=0\\ 4x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ 4x=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac34\end{array}\right.\)
b: \(2x^2-4x+2=0\)
=>\(2\left(x^2-2x+1\right)=0\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
c: \(x^3-4x^2-x+4=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>(x-4)(x-1)(x+1)=0
=>\(\left[\begin{array}{l}x-4=0\\ x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=1\\ x=-1\end{array}\right.\)
Bài 1:
a: \(2xy+4x^2y^2+2x^3y^3\)
\(=2xy\cdot x^2y^2+2xy\cdot2xy+2xy\cdot1\)
\(=2xy\left(x^2y^2+2xy+1\right)=2xy\left(xy+1\right)^2\)
b:Sửa đề: \(5x^3-10x^2-4x+8\)
\(=\left(5x^3-10x^2\right)-\left(4x-8\right)\)
\(=5x^2\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(5x^2-4\right)\)
c: \(x^2-4y^2+6x+9\)
\(=\left(x^2+6x+9\right)-4y^2\)
\(=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
d: \(4x^2-y^2+6x-3y\)
=(2x-y)(2x+y)+3(2x-y)
=(2x-y)(2x+y+3)
e: \(x^2-9y^2+2x+1\)
\(=\left(x^2+2x+1\right)-9y^2\)
\(=\left(x+1\right)^2-\left(3y\right)^2=\left(x+1-3y\right)\left(x+1+3y\right)\)
