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\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left[\left(x-2\right)\left(x-5\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
Đặt: \(x^2-7x+11=t\)
\(\Rightarrow\hept{\begin{cases}x^2-7x+10=t-1\\x^2-7x+12=t+1\end{cases}}\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
\(=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\)
\(=t^2\)
Vậy: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x^2-7x+11\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 12x. b) 4xy
c) 2y(3 x 2 + y 2 ).
d) (x + y + z)( x 2 + y 2 + z 2 – xy – xz - yz).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x+y)2-(x-y)2
=(x+y)(x-y)
b)(3x+1)2-(x+1)2
=[(3x+1)+(x+1)].[(3x+1)-(x+1)]
=(3x+1+x+1)(3x+1-x-1)
1/ ( x+y)^2 -( x-y)^2 = (x+y-x+y)(x+y+x-y)
=4xy
2/( 3x+1)^2-(x+1)^2= (3x+1-x-1)(3x+1+x+1)
=2x(4x+2)
3/