bằng phương thức nào bạn ?

câu trả lời:

3x.x-13.x+6²

bằng phương pháp đặt nhân tử chung

\(12y-9x^2+36-3x^2y\)

\(=\left(36-9x^2\right)-\left(3x^2y-12y\right)\)

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

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

\(12y-9x^2+36-3x^2y\)

\(=\left(12y-3x^2y\right)-\left(9x^2-36\right)\)

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

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

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

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

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

=(−3)(x−2)(x+2)(3+y)

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

a. \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b. \(=\left(1-x\right)^3\)

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

=(x-y)(x^2+xy+y^2)-(x+y)(x-y)

=(x-y)(x^2+xy+y^2-x-y)

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

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

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

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

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

        \(x^4+3x^2+36\)

\(=\left(x^2\right)^2+2.x^2.6+6^2-9x^2\)

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

      \(2x^4-3x^3-7x^2+6x+8\)

\(=2x^4+2x^3-5x^3-5x^2-2x^2-2x+8x+8\)

\(=2x^3\left(x+1\right)-5x^2\left(x+1\right)-2x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)

\(=\left(x+1\right)\left[2x^2\left(x-2\right)-x\left(x-2\right)-4\left(x-2\right)\right]\)

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

Chúc bạn học tốt.

a/ \(4x^2-9\)

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

b/ \(3x\left(3x-2\right)+1\)

\(=9x^2-6x+1\)

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

\(a,=\left(2x-3\right)\left(2x+3\right)\)

\(b,=9x^2-6x+1=\left(3x-1\right)^2\)

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

\(x^3+3x^2+3x+2\)

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

\(x^3+3x^2+3x+2\)

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

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

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