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\(\left(3x+1\right)^2-\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)
\(=2\cdot6x\)
\(=12x\)
_________
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\cdot\left(x^2+3y^2\right)\)
______
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)
a) Đề bài phải là : \(\left(x+y\right)^2-\left(x-y\right)^2\)thì mới phân tích được.
Nếu đề bài như trên ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2=\)\(\left(x+y-x+y\right)\left(x+y+x-y\right)=2x\cdot2y=4xy\)
b) Ta có: \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
= \(2x\cdot\left(4x+2\right)=2x\cdot2\cdot\left(2x+1\right)=4x\cdot\left(2x+1\right)\)
c) Ta có : \(x^3+y^3+z^3-3xyz\)
= \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xy\)
=\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y+z\right)\)
=\(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
=\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(a,\Rightarrow\left(x-3-5+2x\right)\left(x-3+5-2x\right)=0\\ \Rightarrow\left(3x-8\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\\ b,=\left(x+y\right)^2-\left(x-2y\right)^2\\ =\left(x+y-x+2y\right)\left(x+y+x-2y\right)=3y\left(2x-y\right)\\ c,=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
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/
a) 12x. b) 4xy
c) 2y(3 x 2 + y 2 ).
d) (x + y + z)( x 2 + y 2 + z 2 – xy – xz - yz).
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)