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\(3a^2-6ab+3b^2-12c^2\)

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

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

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

3a² - 6ab + 3b² - 12c² = 3.(a² - 2ab + b² - 4c²) = 3.[(a - b)² - 4c²] = 3.(a - b - 2c).(a - b + 2c)

\(\left(x-3\right)^2=25\)

\(\Rightarrow\left(x-3\right)^2=5^2\)

\(\Rightarrow x-3=5\)

\(\Rightarrow x=8\)

(x - 3)² = 25

(x - 3)² = 5² = (-5)²

=> x - 3 thuộc {5 ; -5}

=> x thuộc {8 ; -2}

Vậy x thuộc {8 ; -2}

(x-3)²=25

x-3=5 (vì 25=5²) hoặc x-3=-5 (vì 25=(-5)²

x=8 hoặc x=-2

a) 3a²-6ab+3b²-12c²

=3.(a²-2ab+b²-4c²)

=3.[(a-b)²-4c²]

=3.(a-b-2c)(a-b+2c)

\(3a^2-6ab-3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=\left(\sqrt{3}\left(a-b\right)\right)^2-\left(\sqrt{12}c\right)^2=\left(\sqrt{3}a-\sqrt{3}b-\sqrt{12}c\right)\left(\sqrt{3}a-\sqrt{3}b+\sqrt{12}c\right)\)

Sửa đề :\(3a^2-6ab+3b^2-12c^2\)=\(3\left(a^2-2ab+b^2-4c^2\right)\)

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

3a² - 6ab + 3b² - 12c²

⁼ 3a² - 3a x 2b + 3b² - 3 x 4c²

= 3(a² -a x b +b² - 4c²)

phân tích thành nhân tử :
3a² - 6ab + 3b² - 12c²

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

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

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

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

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

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

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