Bài 1: Phân tích các đa thức sau thành nhân tử:
a)x2-2x
b) 3y3+6xy2+3x2y
c) x2-2xy-xy+2y2
Bài 2:Phân tích các đa thức sau thành nhân tử
a)x2-y2+5x-5y
b)x2+4x-y2+4
c)x2-6xy+9y2-16
d) 2x2-4x+2-2y2
e)x2+5x+6
g) x4+4
Bài 3: Tìm x biết
a) x3-2x=0
b) x(x-4)+(x-4)=0
c) x(x-3)+4x-12=0
d) x2(x-1)-4x+4=0
e) (5x-4)2-16=0
g)(2x-1)2-(x+3)2=0
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
1.a)\(x^2-2x=x\left(x-2\right)\)
b)\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
c)\(x^2-2xy-xy+2y^2=x\left(x-y\right)-2y\left(x-y\right)=\left(x-2y\right)\left(x-y\right)\)
3.a)\(x^3-2x=0\)
\(\Leftrightarrow x^2\left(x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b)\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
c)\(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)
d)\(x^2\left(x-1\right)-4x+4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x-1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=0\\x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=2\\x=1\end{matrix}\right.\)
e)\(\left(5x-4\right)^2-16=0\)
\(\Leftrightarrow\left(5x-4+4\right)\left(5x-4-4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x=0\\5x-8=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{8}{5}\end{matrix}\right.\)
g)\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Bài 1: Phân tích các đa thức sau thành nhân tử:
a)x2-2x = \(x.\left(x-2\right)\)
b) 3y3+6xy2+3x2y = \(3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
c) x2-2xy-xy+2y2 = \(\left(x^2-2xy\right)-\left(xy-2y^2\right)\)= \(x.\left(x-2y\right)-y.\left(x-2y\right)=\left(x-y\right).\left(x-2y\right)\)
Bài 2:Phân tích các đa thức sau thành nhân tử
a)x2-y2+5x-5y = \(\left(x^2-y^2\right)+\left(5x-5y\right)\)= \(\left(x-y\right).\left(x+y\right)+5\left(x-y\right)=\left(x-y\right).\left(x+y+5\right)\)
b)x2+4x-y2+4=\(\left(x^2+4x+4\right)-y^2\)=\(\left(x+2\right)^2-y^2=\left(x+2-y\right).\left(x+2+y\right)\)
c)x2-6xy+9y2-16=\(\left(x-3y\right)^2-4^2=\left(x-3y-4\right).\left(x-3y+4\right)\)
d) 2x2-4x+2-2y2= \(2.\left(x^2-2x+1-y^2\right)=2.\left[\left(x-1\right)^2-y^2\right]\)
= \(2.\left(x-1-y\right).\left(x-1+y\right)\)
e)x2+5x+6= \(x^2+4x+x+4+2=\left(x^2+4x+4\right)+\left(x+2\right)\)
= \(\left(x+2\right)^2+\left(x+2\right)=\left(x+2\right).\left(x+2+1\right)=\left(x+2\right).\left(x+3\right)\)
g) x4+4 = \(\left(x^2\right)^2+2^2=\left(x^2-2\right).\left(x^2+2\right)\)
Bài 3: Tìm x biết
a) x3-2x=0
=> \(x.\left(x^2-2\right)=0=>x=0\) hay \(x^2-2=0\)=> \(x^2=2=>\)Không tìm đuợc x
Vậy x = 0
b) x(x-4)+(x-4)=0
=> \(\left(x-4\right).\left(x+1\right)=0=>x-4=0\) hay \(x+1=0\)
=> \(x=4\) hay \(x=-1\)
Vậy x = 4 hay x = -1
c) x(x-3)+4x-12=0
\(=>x.\left(x-3\right)+4x-12\)
=> \(x.\left(x-3\right)+4\left(x-3\right)=0\)
=> \(\left(x+4\right).\left(x-3\right)=0\)
=> \(x+4=0\) hay \(x-3=0\)
=> \(x=-4\) hay \(x=3\)
d) x2(x-1)-4x+4=0
=> \(x^2\left(x-1\right)-\left(4x-4\right)=0\)
=> \(x^2\left(x-1\right)-4\left(x-1\right)=0\)
=> \(\left(x^2-4\right)\left(x-1\right)=0\)
=> \(x^2-4=0\) hay \(x-1=0\)
=> \(x^2=4\) hay \(x=1\)
=> \(x=2\) hay \(x=-2\) hay \(x=1\)
Vậy x = 2 hay x = -2 hay x = 1
e) (5x-4)2-16=0
=> \(\left(5x-4\right)^2-4^2=0\)
=> \(\left(5x-4-4\right).\left(5x-4+4\right)=\left(5x-8\right).5x=0\)
=> \(5x-8=0\) hay \(5x=0\)
=> \(5x=8=>x=\dfrac{8}{5}\) hay x = 0
Vậy \(x=\dfrac{8}{5}\) hay x = 0
g)(2x-1)2-(x+3)2=0
=> \(\left(2x-1-x-3\right).\left(2x-1+x+3\right)=0\)
=> \(\left(x-4\right).2=0=>x-4=0=>x=4\)
Vậy x = 4
2.a)\(x^2-y^2+5x-5y=\left(x+y\right)\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(x+y+5\right)\)
b)\(x^2+4x-y^2+4=\left(x+2\right)^2-y^2=\left(x+2+y\right)\left(x+2-y\right)\)
c)\(x^2-6xy+9y^2-16=\left(x-3y\right)^2-4^2=\left(x-3y+4\right)\left(x-3y-4\right)\)
d)\(2x^2-4x+2-y^2=2\left[\left(x-1\right)^2-y^2\right]=2\left(x-1+y\right)\left(x-1-y\right)\)
e)Sai đề: \(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
g)\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
