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

\(=\left(6+2\right)\left(3^2+1\right)\left(3^4+1\right)\)

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

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

\(=3^8-1\)

a) (x + 3)² - 2(x + 3)(x - 2) + (x - 2)² 

= (x + 3 - x + 2)² = 5² = 25

b) (2x + 5)² + 2(2x + 5)(3x - 1) + (3x - 1)² 

= (2x  + 5 + 3x - 1)² = (5x  + 4)²

Trả lời:

a/ ( x + 3 )² - 2 ( x + 3 ) ( x - 2 ) + ( x - 2 )²

=  [ ( x + 3 ) - ( x - 2 ) ]² 

= ( x + 3 - x + 2 )²

= 5²

= 25

b/ ( 2x + 5 )² + 2 ( 2x + 5 ) ( 3x - 1 ) + ( 3x - 1 )²

= ( 2x + 5 + 3x - 1 )²

= ( 5x + 4 )²

a) \(\left(2x+1\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)

\(=8x^3+12x^2+6x+1\)

b) \(\left(x-3\right)^3\)

\(=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

Bài 2: 

a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)

b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)

c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)

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

cái trên của bạn có sai không vậy ?? hình như chỗ -2x phải là -12x

8-12x+6x-x³ =(2-x)³

8-12x+6x²-x³=2³-3.2²x+3.2.x²-x³=(2-x)³

^ok?

giải : 8-2x+6x² -x³ =(2-x)³

      bạn gửi mấy câu khác đi

Bài 1: 

c: \(\left(-5x-y\right)^3=-125x^3-75x^2y-15xy^2-y^3\)

h: \(\left(3y-2x^2\right)^3=27y^3-54y^2x^2+36yx^4-8x^6\)

Bài 1:

c. (-5x - y)³ = -125x³ - 50x²y - 10xy² - y³

d. (3y - 2x²)³ = 27y³ - 18x²y² + 24xy⁴ - 8x⁶

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

\(\dfrac{9}{16}-\dfrac{1}{4}x^2=\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)\left(\dfrac{3}{4}+\dfrac{1}{2}x\right)\)

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