x2 hay x^2

là ^2 có dg ko

a) 16 + 2x³y³

= 2( 8 + x³y³)

= 2[ 2³ + (xy)³]

= 2 (2+xy)(4 - 2xy + x²y²)

b) 100a² - (a² + 25)²

= (10a)² - (a² +25)²

= (10a - a² - 25)(10a + a² +25)

= -(a² - 2a.5 + 5²)(a² + 2a.5 + 5²)

=-(a-5)² (a+5)²

a) \(16+2x^3y^3\)

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

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

b) \(100a^2-\left(a^2+25\right)^2\)

\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)

\(=-\left(a-5\right)^2\left(a+5\right)^2\)

2x - 2y - x² + 2xy - y²

= (2x - 2y) - (x² - 2xy + y²)

= 2(x - y) - (x - y)²

= (x - y)(2 - x + y)

a: \(3x^3-75x\)

\(=3x\left(x^2-25\right)\)

\(=3x\left(x-5\right)\left(x+5\right)\)

b: \(x^4y^2-12x^3y^2+48x^2y^2-64xy^2\)

\(=xy^2\left(x^3-12x^2+48x-64\right)\)

\(=xy^2\cdot\left(x-4\right)^3\)

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

b) \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2=10xy^2\left(x^3-x^2-x+1\right)=10xy^2\left(x-1\right)^2\left(x+1\right)\)

a: \(x^4-4x^2-4x-1\)

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

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

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

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

b: \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2\)

\(=10xy^2\left(x^3-x^2-x+1\right)\)

\(=10xy^2\cdot\left[\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\right]\)

\(=10xy^2\cdot\left(x+1\right)\left(x-1\right)^2\)

a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)

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

c) \(3x^2y-12y=3y\left(x^2-4\right)=3y\left(x-2\right)\left(x+2\right)\)

d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)

a: \(5x^2y-20xy+20y\)

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

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

b: \(3x^3+6x^2+3x\)

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

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

c: \(3x^2y-12y\)

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

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

d: \(7x^3-28x^2+28x\)

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

\(=7x\left(x-2\right)^2\)

a: \(x^2-4xy+4y^2-2x+4y-35\)

\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)-35\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)

\(=\left(x-2y\right)^2-7\left(x-2y\right)+5\left(x-2y\right)-35\)

\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)

\(=\left(x-2y-7\right)\left(x-2y+5\right)\)

c: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)

\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)

\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)

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

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

\(\left(a+4\right)^2-16a^2\)

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

\(=\left(a+4+4a\right)\left(a+4-4a\right)\)

\(=\left(5a+4\right)\left(4-3a\right)\)

\(x^3=4x\)

\(\Leftrightarrow x^3-4x=0\)

\(\Leftrightarrow x\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{4}=\pm2\end{cases}}\)

c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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

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

c: \(5x^2+15x+3y+xy\)

\(=5x\left(x+3\right)+y\left(x+3\right)\)

\(=\left(x+3\right)\left(5x+y\right)\)

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

\(=\left(x-y\right)^2-9\)

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

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

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

2) \(6x^2+12xy+6y^2=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

3) \(a^3b-ab^3+a^2+2ab+b^2=ab\left(a^2-b^2\right)+\left(a+b\right)^2=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)

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