$2xy(3xy+2xy^2)$

$=2xy\cdot3xy+2xy\cdot2xy^2$

$=6x^2y^2+4x^2y^3$

$(2x-1)(x^2+2x+4)-(x^2-3x)\cdot2x$

$=2x^3+4x^2+8x-x^2-2x-2x^3+6x^2$

$=9x^2+6x$

$4x^3y-8x^2y^2+4xy^3$

$=4xy(x^2-2xy+y^2)$

$=4xy(x-y)^2$

$2xy+3xz+6y^2+xz$

$=2xy+4xz+6y^2$

$=2(xy+2xz+3y^2)$

Biểu thức trong ngoặc không phân tích tiếp được bằng các phương pháp thông thường.

$y^2-4x-4xy+4x^2+2y$

$=y^2+2y+1+4x^2-4xy-4x-1$

$=(y+1)^2+(2x-y)^2-4x-1$

Biểu thức này không có dạng nhân tử đơn giản với hệ số nguyên. Có khả năng đề câu c bị chép sai.

Bài 3:

A(x)⋮B(x)

=>\(3x^2+5x+m\) ⋮x-2

=>\(3x^2-6x+11x-22+m+22\) ⋮x-2

=>m+22=0

=>m=-22

Bài 2:

a: \(2x^3-8x^2+8x\)

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

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

b: 2xy+2x+yz+z

=2x(y+1)+z(y+1)

=(y+1)(2x+z)

c: \(x^2+2x+1-y^2\)

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

=(x+1-y)(x+1+y)

Câu 1:

a:\(\left(4x-1\right)\left(2x^2-x-1\right)\)

\(=8x^3-4x^2-4x-2x^2+x+1\)

\(=8x^3-6x^2-3x+1\)

b: \(\left(4x^3+8x^2-2x\right):2x\)

\(=\frac{4x^3}{2x}+\frac{8x^2}{2x}-\frac{2x}{2x}\)

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

c: \(\left(6x^3-7x^2-16x+12\right):\left(2x+3\right)\)

\(=\left(6x^3+9x^2-16x^2-24x+8x+12\right):\left(2x+3\right)\)

\(=\left\lbrack3x^2\left(2x+3\right)-8x\left(2x+3\right)+4\left(2x+3\right)\right\rbrack:\left(2x+3\right)\)

\(=3x^2-8x+4\)

c) Ta có: \(P=x^3+y^3+6xy\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+6xy\)

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

\(=2^3=8\)

a: \(y^3+2xy^2+y^2-4x^2\)

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

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

\(\frac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\)

\(=\frac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{\left(2x+y\right)\left(y^2+y-2x\right)}=\frac{4x^2-2xy+y^2}{y^2+y-2x}\)

b: \(\frac{x^2-2x-8}{2x^2+9x+10}\)

\(=\frac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

\(=\frac{\left(x-4\right)\cdot\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}=\frac{x-4}{2x+5}\)

c: \(\frac{6x-x^2-5}{5x^6-x^7}\)

\(=\frac{x^2-6x+5}{x^7-5x^6}\)

\(=\frac{\left(x-5\right)\left(x-1\right)}{x^6\cdot\left(x-5\right)}=\frac{x-1}{x^6}\)

d: \(\frac{x^3+64}{2x^3-8x^2+32x}=\frac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}=\frac{x+4}{2x}\)

e: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

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

\(=\frac{x+y}{\left(x-y\right)\left(x+y\right)}=\frac{1}{x-y}\)

câu c (x+3)(x^2-3x+9)-(54+x^3)=x^3+27-54-x^3

=27

a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

c) \(x^3-x^2+2\)

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

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

f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)

g) \(6x^2-12x=6x\left(x-2\right)\)

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

i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

k) \(2x^3+2x^2y-4xy^2=2x\left(x^2+xy-2y^2\right)\)

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)