Bài 1 :

Tự phân tích vế trái và điền vào vế phải 

Bài 2 :

a) \(3x^3-6x^2+3x\)

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

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

b) \(2xy+z+2x+yz\)

\(=\left(2xy+2x\right)+\left(z+yz\right)\)

\(=2x\left(y+1\right)+z\left(y+1\right)\)

\(=\left(y+1\right)\left(2x+z\right)\)

c) \(x^4-y^4\)

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

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

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

d) \(3x^2-4x-7\)

\(=3x^2+3x-7x-7\)

\(=3x\left(x+1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-7\right)\)

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

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

\(=x^3+3x^2y+3xy^2+y^3\)

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

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

\(=x^3-3x^2y+3xy^2-y^3\)

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3-y^3\)

\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

a) \(\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)

b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)

\(=\dfrac{\left(x+y\right)^2}{x^2+xy+x^2-y^2}=\dfrac{\left(x+y\right)^2}{x\left(x+y\right)+\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{x+y}{\left(2x-y\right)}\)

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=ax\left(x-a\right)\)

a, \(\left(\frac{1}{2}xy-\frac{3}{4}\right)\left(\frac{1}{2}xy+\frac{3}{4}\right)=\frac{1}{4}x^2y^2+\frac{3}{8}xy-\frac{3}{8}xy-\frac{9}{16}=\frac{1}{4}x^2y^2-\frac{9}{16}\)

b, \(\left(2x+3\right)\left(4x^2-6x+9\right)=8x^3-12x^2+18x+12x^2-18x+27=8x^2+27\)

c, \(\left(xy-2\right)\left(x^2y^2+2xy+4\right)=x^3y^3+2x^2y^2+4xy-2x^2y^2-4xy-8=x^3y^3-8\)

Mk ko chép đề bài ra nhé

a, = 1/2xy.( -3/4 + 3/4 )

   =1/2xy.

b, Áp dụng HĐT số 2, có:  (Chỗ này ko cần chép cx đc)

=(2x+3).(2x+3)^2  (^ là mũ)

=(2x+3)^3

c, Áp dụng HĐT số 2, có:

=(xy-2).(xy + 2)^2

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

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

b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)

c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)

d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)

\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)

e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

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

f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)

\(=5^2-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)

1,

a, (2x + 1- x + 3)² = (x+4)²

b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)

c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)

d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)

e,\(\left(x+y\right)^2-z\left(x+y\right)\)=\(\left(x+y\right)\left(x+y-z\right)\)

f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)

\(=\left(5+2x+y\right)\left(5-2x-y\right)\)

Chúc các bn hc tốt