\(x^4+13x^2+36=x^4+4x^2+9x^2+36\)

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

       \(x^4+13x^2+36\)

<=> \(x^4+9x^2+4x^2+36\)

<=> \(x^2\left(x^2+9\right)+4\left(x^2+9\right)\)

<=> \(\left(x^2+9\right)\left(x^2+4\right)\)

\(3x^3-4x^2+13x-4\)

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

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

a) \(36-4x^2+4xy-y^2\)

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

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

b) \(2x^4+3x^2-5\)

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

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

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

thank bn

\(=2x^4-6x^3-x^3+3x^2-5x^2+15x-2x+6\)

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

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

\(=\left(x-3\right)\left(2x^3-4x^2+3x^2-6x+x-2\right)\)

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

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

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

=x³(x+2)-13x²+12x-26x+24

=x³(x+2)-x(13x-12)-2(13x-12)

=x³(x+2)-(13x-12)(x+2)

=(x+2)(x³-x-12x+12)

(x+2)[(x²-1)-12(x-1)]

=(x+2)[x(x-1)(x+1)-12(x-1)]

=(x+2)(x-1)[x(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

=(x+2)(x-1)[x(x-3)+4(x+3)]

=(x+2)(x-1)(x-3)(x+4)

trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!

Bài làm:

a) \(x^2-6x+4=\left(x^2-6x+9\right)-5=\left(x-3\right)^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)

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

c) \(6x^2-5x+1=6x^2-3x-2x+1=\left(2x-1\right)\left(3x-1\right)\)

d) \(3x^2+13x-10=3x^2+15x-2x-10=\left(x-5\right)\left(3x-2\right)\)

\(4x^3-13x^2+9x-18=4x^3-12x^2-x^2+3x+6x-18\)

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

\(4x^3-13x^2+9x-18\)

\(=4x^3-12x^2-x^2+3x+6x-18\)

\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(4x^2-x+6\right)\)

\(4x^3-13x^2+9x-18 \)

\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(4x^2-x+6\right)\)