\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)

x²-7x+12

=x²-4x-3x+12

=x(x-4)-3(x-4)

=(x-3)(x-4)

        \(x^2-7x+12\)

\(=\left(x^2-4x\right)-\left(3x-12\right)\)

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

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

c ) \(x^2-7x+12\)

\(=\left(x^2-3x\right)-\left(4x-12\right)\)

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

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

d ) \(x^2+7x+12\)

\(=\left(x^2+3x\right)+\left(4x+12\right)\)

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

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

Phân tích đa thức thành nhân tử:

a) x² (2x - 5) + 6x - 15 =  x² (2x - 5) + 3(2x - 5) =( x2 + 3 )(2x - 5)

b)x² + 7x + 12 =  x² + 3x + 4x + 12 = (x + 3 )( x + 4)

\(12x^2+7x-12\)

\(=12x^2-9x+16x-12\)

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

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

12x^2+7x-12=(12x^2-9x)+(16x-12)=3x(4x-3)+4(4x-3)=(3x+4)(4x-3)

chúc hok tốt

Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với  - Hoc24

a) \(x^2+5x+4==x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

b) \(3x^2+4x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(x-1\right)\left(3x+7\right)\)

c) \(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

a) (x+1)(x+4)

b) (x-1)(3x+7)

c) (x+3)(x+4)

a) x²+5x+4 = x(x+4)+(x+4) = (x+4)(x+1)

b) 3x²+4x-7 = 3x(x-1)+7(x-1) = (x-1)(3x+7)

c) x²+7x+12 = x(x+4)+3(x+4) = (x+3)(x+4)

\(x^2+7x+12\)

cách 1: \(=x^2+4x+3x+12\)

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

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

cách 2: \(=x^2+3x+4x+12\)

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

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

cách 3: \(=\left(x^2+7x+12,25\right)-0.25\)

\(=\left(x+3.5\right)^2-0.5^2\)

\(=\left(x+3.5+0.5\right)\left(x+3.5-0.5\right)\)

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

lấy đâu ra 8 cách vậy trời!!!!!!!!!!!!!!!

Cách 1: 

   \(x^2+7x+12\)

\(=\left(x^2+4x\right)+\left(3x+12\right)\)

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

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