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)

\(\left(3x+1\right)^2-\left(3x-1\right)^2\)

\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

\(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\cdot\left(x^2+3y^2\right)\)

______

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)

a) 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) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)

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) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)

hằng đẳng thức a²-b²=(a-b)(a+b) í bạn

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/ \(=x^4+x^3+x^2+5x^2+5x+5\)

\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)=\left(x^2+5\right)\left(x^2+x+1\right)\)

b/ \(=x^3+x^2+2x-x^2-x-2\)

\(=x\left(x^2+x+2\right)-\left(x^2+x+2\right)=\left(x-1\right)\left(x^2+x+2\right)\)

c/ \(=x^3+4x^2+4x-x^2-4x-4\)

\(=x\left(x^2+4x+4\right)-\left(x^2+4x+4\right)=\left(x-1\right)\left(x+2\right)^2\)

câu d khó quá , mk lm k nổi , sr nha ^^

a) x⁴ + x³ + 6x² + 5x + 5
= x⁴ + x³ + x² + 5x² + 5x + 5
= x² ( x² + x + 1) + 5 (x² + x + 1)
= (x² + x + 1) (x² + 5)

b) x³ + x - 2
= x³ + x² + 2x - x² - x - 2
= x (x² + x + 2) - (x² + x + 2)
= (x² + x + 2) (x - 1)

c) x³ + 3x² - 4
= x³ + 4x² + 4x - x² - 4x - 4
= x (x² + 4x + 4) - (x² + 4x + 4)
= (x² + 4x + 4) (x - 1)
= (x + 2)² (x - 1)

d) xy(x + y) + yz(y + z) + xz(x + z) + 3xyz
= xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + y + z)
= (x + y + z) (xy + yz + xz)

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)