a, 15x³y⁵z : 5x²y³ = 3xy²z.

b, 12x⁴y² : ( - 9xy² ) = \(\frac{3}{4}x^3\).

c, ( 30x⁴y³ - 25x²y³ - 3x⁴y⁴ ) : 5x²y³ = \(6x^2-5-\frac{3}{5}x^2y.\)

d, ( 4x⁴ - 8x²y² + 12x⁵y ) : ( - 4x² ) = -x² + 2y² - 3x³y.

a: \(=\dfrac{15}{5}\cdot\dfrac{x^3}{x^2}\cdot\dfrac{y^5}{y^3}\cdot z=3xy^2z\)

b: \(=-\dfrac{4}{3}x^3\)

c: \(=\dfrac{30x^4y^3}{5x^2y^3}-\dfrac{25x^2y^3}{5x^2y^3}-\dfrac{3x^4y^4}{5x^2y^3}\)

\(=6x^2-5-\dfrac{3}{5}x^2y\)

d: \(=\dfrac{4x^4}{-4x^2}+\dfrac{8x^2y^2}{4x^2}-\dfrac{12x^5y}{4x^2}\)

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

Giúp tui với mấy bạn ơi

C

D

Sửa lại:

a, x² + 6x + 9

a) x² + 6x + 9 = (x + 3)²

b) 25x² - 30x + 9 = (5x - 3)²

c) (2x - 3y)(2x + 3y) = (2x)² - (3y)² = 4x² - 9y²

d) (x + 2y)(x² + 2xy + 4y²) = (x + 2y)³

e) 27x³ + 54x² + 36x + 8

Bài này chủ yếu là dùng HĐT

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

b) \(25x^2-30x+9=\left(5x-3\right)^2\)

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

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

e) \(27x^3+54x^2+36x+8=\left(3x\right)^3+3.\left(3x\right)^2.2+3.3x.2^2+2^3=\left(3x+2\right)^3\)

f) \(\left(3x-5y\right)\left(9x^2+15xy+25y^2\right)=\left(3x-5y\right)\left[\left(3x\right)^2+3x.5y+\left(5y\right)^2\right]=\left(3x\right)^3-\left(5y\right)^3=27x^3-125y^3\)

g) \(x^3-15x^2+75x-125=x^3-3.x^2.5+3.x.5^2-5^3=\left(x-5\right)^3\)

a) (4x-1)(2-x)-(2x-1)²

= 8x-4x²-2+x-(4x²-4x+1) = -8x²+13x-3

b) (15x⁴y⁵-30x³y⁴+35x³y⁴):(5x³y³)

= 3xy²-6y+7y = 3xy²+y

a: \(=8x-4x^2-2+2x-4x^2+4x-1\)

\(=-8x^2+14x-3\)

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

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

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

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

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

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

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=5x^2y^2\left(x-1\right)^2\)

3) \(12x^2-12xy+3y^2\)

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

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

4) \(8x^3-8x^2y+2xy^2\)

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

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

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

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

1: 4x^5y^2-8x^4y^2+4x^3y^2

=4x^3y^2(x^2-2x+1)

=4x^3y^2(x-1)^2

2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)

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

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

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

a) (25x^5 – 5x^4 + 10x^2) : 5x^2

= (25x^5 : 5x^2) – (5x^4 : 5x^2) + (10x^2 : 5x^2)

= 5x^3 – x^2 + 2

b) (15x^3y^2- 6x^2y – 3x^2y^2) : 6x^2y

= (15x^3y^2 : 6x^2y) + (-6x^2y : 6x^2y) + (- 3x^2y^2 : 6x^2y)

=\(\frac{15}{6}xy-1-\frac{3}{6}y\)

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

Xét:

\(\left(3x-2y\right)\left(25x^2-9y^2\right)\)

\(=\left(3x-2y\right)\left(5x-3y\right)\left(5x+3y\right)\)

\(=\left(5x-3y\right)\left(15x^2+9xy-10xy-6y^2\right)\)

\(=\left(5x-3y\right)\left(15x^2-xy-6y^2\right)\)

Từ đó dễ dàng suy ra tích chéo = nhau => đpcm

ta có : \(VP=\dfrac{15x^2-xy-6y^2}{25x^2-9y^2}=\dfrac{\left(3x-2y\right)\left(5x+3y\right)}{\left(5x-3y\right)\left(5x+3y\right)}=\dfrac{3x-2y}{5x-3y}=VT\)

a: =-2/5x^5y^7

Hệ số: -2/5

bậc: 12

b: =3/4*x^2y^3*12/5x^4=9/5x^6y^3

Hệ số: 9/5

bậc: 9

c: =4/9x^6y^6

hệ số: 4/9

bậc: 12

d: =2/5x^6y^6

hệ số: 2/5

bậc: 12