(4x2 – 9y2) : (2x – 3y)

(Sử dụng HĐT để phân tích số bị chia thành tích)

= [(2x)2 – (3y)2] : (2x – 3y)

(Xuất hiện hằng đẳng thức (3))

= (2x – 3y)(2x + 3y) : (2x – 3y)

= 2x + 3y.

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

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)

(15³y²-6x²y-3x²y²):6x²y

=\(\dfrac{5}{2}\)xy-1-\(\dfrac{1}{2}\)y

(4x²-9y²):(2x-3y)

=(2x-3y)(2x+3y):(2x-3y)

=2x+3y

\(a,=2xy\left(\dfrac{1}{3}x-y+2\right)\\ b,=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)=\left(2x+3y\right)\left(2x-3y-1\right)\)

b: Ta có: \(4x^2-9y^2-2x-3y\)

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

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

\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)

\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

a ) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)\)

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

\(=2x+y\)

b ) \(\left(27x^3+1\right):\left(3x+1\right)\)

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

\(=9x^2-3x+1\)

c ) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)\)

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

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

\(=3y-x\)

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

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

\(=2x-1\)

:D

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

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

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

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

$=[(3x+2)-(2y-1)]^2$

$=(3x-2y+3)^2$

$=(x^2+2xy)^2+2(x^2+2xy)y^2+(y^2)^2$

$=(x^2+2xy+y^2)^2$

d) $(x-1)^3+3x(x-1)^2+3x^2(x-1)+x^3$

Đặt $a=x-1,\ b=x$:

$=a^3+3a^2b+3ab^2+b^3$

$=(a+b)^3$

$=[(x-1)+x]^3$

$=(2x-1)^3$

Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:

$=(2x)^3+(3y)^3$

$=8x^3+27y^3$

$=x^3-y^3-(x^3+y^3)$

$=-2y^3$

Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:

$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$

$=x^6-8y^3-x^6+x^3y^3+8y^3$

$=x^3y^3$

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

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

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

a) $(2x-1)^2-2(2x-3)^2+4$

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

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

$=[(3x+2)-(2y-1)]^2$

$=(3x-2y+3)^2$

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

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

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

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

a) x² - 10x + 25x = ( x - 5)²

b) 4 - 4x² + x⁴ = ( 2 - x² )²

c) x² - 6xy + 9y² = (x - 3y )²

d_ (2x + y² ). (2x - y² ) = 4x² - y⁴