ban con choi bangbang ko cho minh muon nick

`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.

`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.

`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.

\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a) 15x²-5x³=5x²(3-x)

a: \(15x^2-5x^3=5x^2\left(3-x\right)\)

b: \(8x^3-y^3+4x^2y-2xy^2\)

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

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

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

c: Ta có: \(x^8+64y^4\)

\(=x^8+16x^4y^2+64y^4-16x^4y^2\)

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

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

A = (x^2 - y^2 + 2x) + 1 = (x^2 + 2x + 1)-y^2 = (x+1)^2 - y^2 =

(x+1-y)(x+1+y)

B = x^4 - 64y^4 = (x^2)^2 - (8y^2)^2 = (x^2 - 8y^2 ) (x^2 + 8y^2 )

C = (x^2 + x)^2 + 2(x^2 + x) - 3

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

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

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

D = (x^2 - 4x)^2 + 7(x^2 - 4x) + 12

= (x^2 - 4x)^2 + 2(x^2 - 4x). 7/2 + 49/4 - 1/4

= (x^2 - 4x + 7/2)^2 - 1/4

= (x^2 - 4x + 7/2 - 1/4)(x^2 - 4x + 7/2 + 1/4)

= (x^2 - 4x + 13/4)(x^2 - 4x +15/4)

D = (x^2 - 4x)^2 + 7(x^2 - 4x) + 12

= (x^2 - 4x)^2 + 2(x^2 - 4x).7/2 + 49/4 - 1/4

= (x^2 - 4x + 7/2)^2 - 1/4

= (x^2 - 4x + 7/2 - 1/2)(x^2 - 4x + 7/2 + 1/2)

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

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

\(9x^2-30xy+25y^2=\left(3x\right)^2-2\cdot3x\cdot5y+\left(5y\right)^2=\left(3x-5y\right)^2\)

\(4x^2+16x+7=\left(2x^2\right)+2\cdot2x\cdot4+4^2-9=\left(2x+4\right)^2-3^2=\left(2x+1\right)\left(2x+7\right)\)

\(-5+18y-9y^2=-\left[\left(3y\right)^2-2\cdot3y\cdot3+3^2-4\right]=-\left[\left(3y-3\right)^2-2^2\right]=-\left(3y-5\right)\left(3y-1\right)\)

đến đó giải tiếp hộ mik nx ddi đc k

a: \(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)=\left(2x+3y\right)^2:\left(2x+3y\right)=2x+3y\)

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

e: \(\dfrac{64y^3-27}{4y-3}=\dfrac{\left(4y-3\right)\left(16y^2+12y+9\right)}{4y-3}=16y^2+12y+9\)

a, \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)

\(\Rightarrow\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)

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

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

\(\Rightarrow\left(x^2+6xy+9y^2\right):\left(x+3y\right)\\ =\left(x+3y\right)^2:\left(x+3y\right)\\ =x+3y\)

c, \(64y^3-27=\left(4y-3\right)\left(16y^2+12y+9\right)\)

\(\Rightarrow\left(64x^3-27\right):\left(4y-3\right)\\ =\left[\left(4y-3\right)\left(16x^2+12x+9\right)\right]:\left(4y-3\right)\\ =16x^2+12x+9\)

1) \(=\left(2z+3\right)\left(4z^2-6z+9\right)\)

2) \(=\left(\frac{3x^2}{5}-\frac{1}{2}\right)\left(\frac{3x^2}{5}+\frac{1}{2}\right)\)

3) \(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

4) \(=\left(2x+1\right)^2\)

5) \(=\left(x-10\right)^2\)

6) \(=\left(y^2-7\right)^2\)

7) \(=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

Cảm ơn bạn nhiều nha 😁😁😁😁

a/ \(=64y^4+32xy^3+8y^2x^2-32xy^3-16x^2y^2-4x^3y+8x^2y^2+4x^3y+x^4\)

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

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

b/ \(=y^4+2xy^3+2x^2y^2-2xy^3-4x^2y^2-4x^3y+2x^2y^2+4x^3y+4x^4\)

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

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

c/ \(=x^4+5x^3+7x^2+5x^3+25x^2+35x+3x^2+15x+21\)

\(=x^2\left(x^2+5x+7\right)+5x\left(x^2+5x+7\right)+3\left(x^2+5x+7\right)\)

\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)

d/ \(=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)

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

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

1) \(x^2+6x+9\)

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

2) \(10x-25-x^2\)

\(=-25+10x-x^2\)

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

3) \(8x^3-\dfrac{1}{8}\)

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

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

4) \(\dfrac{1}{25}x^2-64y^2\)

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

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

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

\(10x-25-x^2=-\left(x-5\right)^2\)

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