\(3a^2c^2+bd+3abc+acd\)

\(=\left(3a^2c^2+3abc\right)+\left(acd+bd\right)\)

\(=3ac\left(ac+b\right)+d\left(ac+b\right)\)

\(=\left(ac+b\right)\left(3ac+d\right)\)

a) 25 - x² + 4xy - 4y² = 25 - (x² - 4xy + 4y²) = 5² - (x - 2y)² = (5 + x - 2y)(5 - x +2y) = (x - 2y + 5)(2y - x + 5)

b) 3a²c² + bd + 3abc + acd = (3a²c²​ + 3abc) + (bd + acd) = 3ac(ac + b) + d (ac + b) = (ac + b)(3ac + d)

c) x³ - 2x² - x + 2 = x²(x - 2) - (x - 2) = (x - 2)(x² - 1) = (x - 2)(x - 1)(x + 1)

d) a⁴ + 5a³ + 15a - 9 = (a⁴ + 3a²) + (5a³ + 15a) - (3a² + 9) = a²(a² + 3) + 5a(a² + 3) - 3(a² + 3) = (a² + 3)(a² + 5a - 3)

4x^2=28x+49

4x^2+28X+49

 Thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

Biến đổi vế trái thành: 

a^3+b^3+c^3-3abc 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc 

<=>[(a+b)^3 +c^3] -3ab.(a+b+c) 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c) 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)

boc vai

Ta có

a³+b³+c³-3abc

=(a+b)³-3ab(a+b)+c³-3abc

=[(a+b)³+c³]-3ab(a+b+c)

=(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)

=(a=b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c)

=(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)

=(a+b+c)(a²+b²+c²-ab-ac-bc)

a³+b³+c³-3abc

=(a+b)³-3ab(a+b)+c³-3abc

=[(a+b)³+c³]-3ab(a+b+c)

=(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)

=(a=b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c)

=(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)

=(a+b+c)(a²+b²+c²-ab-ac-bc)

a³ + b³ + c³ - 3abc

= [( a³ + b³ ) + c³ ] - 3abc

= [( a + b )³ + c³ + 3ab( a + b )] - 3abc

= [( a + b )³ + c³ ] + 3a²b + 3ab² - 3abc

= ( a + b + c ) [( a + b )² - c( a + b ) + c² ] - 3ab( a + b + c )

= ( a + b + c ) ( a² + 2ab + b² - ac - bc + c² - 3ab )

= ( a + b + c ) ( a² + b² + c² - ab - bc - ac )

= 1/2( a + b + c ) ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ac )

= 1/2( a + b + c ) [( a² - 2ab + b² ) + ( a² - 2ac  + c² ) + ( b² - 2bc + c² )]

= 1/2 ( a + b + c ) [( a - b )² + ( a - c )² + ( b - c )²]

Hok Tốt!!!

\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)

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

\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)

\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

Ta Có : \(a^3+b^3+c^3-3abc \)

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

\(=a^3+3a^2b+ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b\right)^3+c^3-3abc\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

= \(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2-ab+b^2-ac-bc+c^2\right)\)

 ta có: ab(a + b) + bc(b + c) + ac(a + c) + 3abc 

= ab(a + b) + abc + bc(b + c) + abc + ac(a + c) + abc 

= ab(a + b + c) + bc(a + b + c) + ac(a + b + c) 

= (a + b + c)(ab + bc + ca) 

\(a^3+b^3-c^3+3abc\)

\(=a^3+3ab.\left(a+b\right)+b^3-c^3-3abc-3ab.\left(a+b\right)\)

\(=\left(a+b\right)^3+c^3-3ab.\left(a+b-c\right)\)

\(=\left(a+b+c\right).\left(a^2+ab+b^2-ab-ac+c^2\right)-3ab.\left(a+b+c\right)\)

\(=\left(a+b+c\right).\left(a^2+b^2+c^2-ab-bc-ca\right)\)