a) 15x + 15y = 15(x + y)

b) 6x - 10y = 2(3x - 5y)

c) 2a + 4b - 6c = 2(a + 2b - 3c)

d) 6xy - 12x - 18y = 6(xy - 2x - 3y)

e) 2(x + y) - 5a(x + y) = (2 - 5a)(x + y)

f) 6x(x - y) + 5(y - x) = 6x(x - y) + (-5)(x - y) = (6x - 5)(x - y)

a) \(15x+15y=15\left(x+y\right)\)

b) \(6x-10y=2\left(3x-5y\right)\)

c) \(2a+4b-6c=2\left(a+2b-3c\right)\)

d) \(6xy-12x-18y=6\left(xy-2x-3y\right)\)

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

f) \(6x\left(x-y\right)+5\left(y-x\right)=6x\left(x-y\right)-5\left(x-y\right)=\left(6x-5\right)\left(x-y\right)\)

a, \(15x+15y=15\left(x+y\right)\)

b, \(6x-10y=2\left(3x-5y\right)\)

c, \(2a+4b-6c=2\left(a+2b-3c\right)\)

d, \(6xy-12x-18y=2\left(3xy-6x-9y\right)\)

e, \(2\left(x+y\right)-5a\left(x+y\right)=\left(x+y\right)\left(2-5a\right)\)

f, \(6x\left(x-y\right)+5\left(y-x\right)=6x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(6x-5\right)\)

full nhé ^^

a)\(2a^2-3ab+b^2\)

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

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

=\(\left(a-b\right)\left(2a-b\right)\)

b)\(x^2-7x-30\)

=\(x^2-10x+3x-30\)

=\(x\left(x-10\right)+3\left(x-10\right)\)

=\(\left(x-10\right)\left(x+3\right)\)

c)\(6a^2-5ab-6b^2\)

=\(6a^2-9ab+4ab-6b^2\)

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

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

d)\(a^4+a^2+1\)

=\(a^4+2a^2-a^2+1\)

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

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

e)\(x^3+6x^2+11x+6\)

=\(x\left(x^2+6x+9+2\right)+6\) 

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

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

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

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

a)   x³ -2x² +5x-4

=x³-x²-x²+x+4x-4

=x²(x-1)-x(x-1)+4(x-1)

=(x²-x+4)(x-1)

b) x³-x²+x+3

=x³+x²-2x²-2x+3x+3

=x²(x+1) -2x(x+1)+3(x+1)

=(x²-2x+3)(x+1)

c) 6x³+x²+x+1

=6x³+ 3x²-2x²-x+2x+1

=6x²(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))

=(6x²-2x+2) (x+\(\frac{1}{2}\))

=2( 3x²-x+1) (x+\(\frac{1}{2}\))

d)  4x³ + 6x²+4x+1

= 4x³+2x²+4x²+2x+2x+1

= 4x²(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))

= 2(2x² +2x+1)( x+\(\frac{1}{2}\))

e) x⁶ -9x³+8

a) \(=3\left(5y+4x\right)\)

b) \(=\left(x-3\right)^2\)

c) \(=y\left(y^2+2y+3\right)\)

\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)

\(=\left(x-3\right)\left(x-1-3\right)\)

\(=\left(x-3\right)\left(x-4\right)\)

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

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

\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

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

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

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

\(=\left(x-5\right)\left(4x+1\right)\)

Còn lại bạn tự làm nhá

Lời giải:

a.

\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)

\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)

\(=-a^4b^4(3a+4b)^2\)

b.

$x^3-6x^2y+12xy^2-8x^3$

$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$

c.

$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$

$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$

$=(x+\frac{1}{2})^3$

a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

\(=-a^4b^4\cdot\left(4b+3a\right)^2\)

b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)

\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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

c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)

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

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

a) Áp dụng HĐT 1 thu được ( 2 x   +   y ) 2 .

b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được

[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).

c) Ta có: 9 - 6x +  x 2  -  y 2 = ( 3   -   x ) 2  -  y 2  = (3 - x - y)(3 -x + y).

d) Ta có: -(x + 2) + 3( x 2  - 4) = -{x + 2) + 3(x + 2)(x - 2)

= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).