bài này 1h rùi,chắc chờ tui ngủ dậy làm;

= (x+y)³ - (x+y) + xy(x+y) =

= (x+y)((x+y)² -1 +xy)) = (x+y)(x² +3xy +y² -1)

Phân tích đa thức thành nhân tử:

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

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

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

Rút gọn biểu thức;

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

\(=\left[\left(6x+1\right)-\left(3x-1\right)\right]^2=\left(6x+1-3x+1\right)=\left(3x+2\right)^2\)

Tìm a để đa thức.. Bạn chia cột dọ thì da

\(xy+y^2-x-y=\left(xy+y^2\right)-\left(x+y\right)=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\)b)\(25-\left(x^2-4xy+4y^2\right)=5^2-\left(x-2y\right)^2=\left(x-2y+5\right)\left(5-x+2y\right)\)

a) x² - 2xy - 4 + y²
= (x - y)² - 2²
= (x - y - 2)(x - y + 2)

b) x² + y² - 1 - 2xy
= (x - y)² - 1²
= (x - y - 1)(x - y + 1)

c) 25 - x² + 4xy - 4y²
= 5² - (x - 2y)²
= (5 - x + 2y)(5 + x - 2y)

\(x^4+x^3+2x^2+x+1\)

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

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

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

x^4+x^3+2x^2+x+1

=(x^4+2x^2+1)+(x^3+x)

=(x^2+1)^2+x(x^2+1)

=(x^2+1)(x^2+x+1)

=1(x⁴+4)

x⁴ + 4 
= x⁴ + 4 + 4x² - 4x² = ( x² + 2 )² - 4x²

= ( x² + 2 - 2x )( x² + 2 + 2x )

P/s: Giang Lê Trà My lần sau giải cho cẩn thận vào nhé.

x²-y²+6x+6y = (x²-y²)+(6x+6y) = (x-y)(x+y)+6(x+y) = (x-y-6)(x+y)

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

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

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~