\(2x^2y^3-\frac{x}{4}-4y^6\)

đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được

\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

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

\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

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

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

Bài 2:

7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)

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

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

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

12) \(\left(6x-1\right)^2-\left(3x+2\right)^2=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)=\left(9x+1\right)\left(3x-3\right)\)

1:

1: ,4x^2-6x=2x(2x-3)

2: 9x^3y^2+3x^2y^2=3x^2y^2(3x+1)

3: x^3+2x^2+3x=x(x^2+2x+3)

4: 2x^2-4x=2x(x-2)

5: 3x-6y=3(x-2y)

6: x^2-3x=x(x-3)

7: 6x^2y+4xy^2+2xy

=2xy(3x+2y+1)

8: 5x^2(x-2y)-15x(x-2y)

=(x-2y)(5x^2-15x)

=5x(x-3)(x-2y)

9: =3(x-y)+5y(x-y)

=(x-y)(5y+3)

10: =(x-1)(3x+5)

11: =2(2x-1)-3(2x-1)

=-(2x-1)

a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)

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

b)  \(64x^3+1=\left(4x+1\right)\left(16x^2-4x+1\right)\)\

c) \(x^3y^6z^9-125=\left(xy^2z^3-5\right)\left(x^2y^4z^6+5xy^2z+25\right)\)

d)  \(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)

e)  \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

64x³ + 1

= ( 4x )³  +  1

= ( 4x + 1 ) ( 16x² - 4x + 1 )

Hằng đẳng thức 6 : A³ + B³

27x⁶ - 8x³

= ( 3x²)³ + ( 2x )³

= ( 3x + 2x ) ( 9x² - 6x + 4x² )

HĐT 6

x⁶ - y⁶

= ( x² )³ - ( y² )³

= ( x² - y² ) ( x⁴ + x²y² + y⁴ )

HĐT 7 : A³ - B³

x³y⁶z⁹ + 1

= ( xy²z³)³ + 1

= ( xy²z³ + 1 ) ( x²y⁴z⁶ + zy²z³ + 1 )

HĐT 6

Ta có : x⁶ - x⁶ 

= ( x²)³  - ( y²)³ 

= ( x² - y² ) ( x⁴ + x²y² + y⁴ ) 

= (x - y)(x + y) ( x⁴ + x²y² + y⁴ ) 

b, <=>(4x)³+1³ 

<=> (4x+1)( 16x²-4x+1)

c, <=> (x.y².z³)³-5³

<=> (xy²z³-5)( x²y⁴z⁶+5xy²z³+25)

d, <=> (3x²)³-(2x)³

<=> (3x²-2x)(9x⁴+6x³+4x²)

d, (x³)²- (y³)² 

= (x³+y³)(x³-y³)

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

4: \(\left(2x+3\right)^3-1\)

\(=\left(2x+3-1\right)\left(4x^2+12x+9+2x+3+1\right)\)

\(=\left(2x+2\right)\left(4x^2+14x+13\right)\)

\(=2\left(x+1\right)\left(4x^2+14x+13\right)\)

5: \(4x^2+20xy+25y^2=\left(2x+5y\right)^2\)

6: \(x^4-64xy^3\)

\(=x\left(x^3-64y^3\right)\)

\(=x\left(x-4y\right)\left(x^2+4xy+16y^2\right)\)