bằng phương thức nào bạn ?

câu trả lời:

3x.x-13.x+6²

bằng phương pháp đặt nhân tử chung

\(y^2-12y+36-144x^2=\left(y-6\right)^2-\left(12x\right)^2=\left(y-12x-6\right)\left(y+12x-6\right)\)

\(=\left(y^2-12y+36\right)-144x^2=\left(y-6\right)^2-144x^2=\left(y-12x-6\right)\left(y+12x-6\right)\)

\(( y − 12 x − 6 ) ( y + 12 x − 6 )\)

   6x^2 + 15x - 36

= 6x^2 + 24x - 9x - 36

= (6x^2 + 24x) - (9x + 36)

= 6x(x + 4) - 9(x+4)

= (6x - 9) (X + 4)

= 3(2x - 3)(x + 4)

\(6x^2+15-36\)

\(=3\left(2x^2+5x-12\right)\)

\(=3\left(2x^2+8x-3x-12\right)\)

\(=3\left[2x\left(x+4\right)-3\left(x+4\right)\right]\)

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

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

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

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

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

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

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

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

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

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

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

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

=(−3)(x−2)(x+2)(3+y)

mình đang cần gấp hộ minh với

\(x^2-36+2xy+y^2=\left(x^2+2xy+y^2\right)-36=\left(x+y\right)^2-6^2=\left(x+y+6\right)\left(x+y-6\right)\)

\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

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

(x²-8)²+36=x⁴-16x²+100=x⁴+20x²+100-36x²=(x²+10)²-36x²=(x²+10-6x)(x²+10+6x)

chuan

Ta có: 

\(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(x^4=16x^2+100\)

$={}^{}-{}^{}$

$=\left({}^{}-6x+10\right)\left({}^{}-6x-10\right)$

=x4−16x2+64+36

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

b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

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

b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)