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

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

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

\(ax^2+ay-bx^2-by=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)

ax^2 + by^2 = 3 chứ không phẢI ax^3 +by^3 = 3 đâu ạ

bn post nhiều nên mình ghi đáp án thôi nhé phần nào sai đề mình cho qua

b)\(\left(x+1\right)\left(xy+1\right)\)

c)\(\left(a+b\right)\left(x+y\right)\)

d)\(\left(x-a\right)\left(x-b\right)\)

e)\(\left(x+y\right)\left(xy-1\right)\)

f)\(\left(a-b\right)\left(x^2+y\right)\)

a) \(x^3-2x^2+2x-1^3\)

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

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

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

b) \(x^2y+xy+x+1\)

\(=xy\left(x+1\right)+\left(x+1\right)\)

\(=\left(xy+1\right)\left(x+1\right)\)

c) \(ax+by+ay+bx\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(a+b\right)\left(x+y\right)\)

d) \(x^2-\left(a+b\right)x+ab\)

\(=x^2-ax-bx+ab\)

\(=\left(x^2-ax\right)-\left(bx-ab\right)\)

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

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

e) Ko biết làm

f) \(ax^2+ay-bx^2-by\)

\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)

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

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

a, x³ - 2x² + 2x - 1³

= x³ - 2x² . 1+ 2x.1² - 1³

= (x - 3 )³

CHTT nha bạn !

\(\left(ax+by\right)^2=1\Leftrightarrow\left(ax\right)^2+2abxy+\left(by\right)^2=1\Leftrightarrow2xy\le1\Leftrightarrow xy\le\frac{1}{2}\)

1/ \(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right).\)

\(=x^2+6x+9-\left(2x^2+6x-5x-15\right)\)

\(=x^2+6x+9-2x^2-6x+5x+15\)

\(=-x^2+5x+24\)

\(=-\left(x^2-5x-24\right)\)

\(=-\left(x^2-8x+3x-24\right)\)

\(=-\left[x\left(x-8\right)+3\left(x-8\right)\right]\)

\(=-\left(x-8\right)\left(x+3\right)\)

2/ \(x^2-xy+x-y\)

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

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

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

3/ \(x^3+6x^2+9x\)

\(=x\left(x^2+6x+9\right)\)

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

1,Bạn tự lm

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

\(3,x^3+6x^2+9x=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\)

\(4,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(xy+1\right)\)

\(5,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(y+x\right)=\left(x+y\right)\left(a+b\right)\)

\(6,x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-bx\right)-\left(ax-ab\right)=x\left(x-b\right)-a\left(x-b\right)=\left(x-b\right)\left(x-a\right)\)