\(a,\)\(x^3-13x-12\)

\(=x^3-x-12x-12\)

\(=x\left(x^2-1\right)-12\left(x+1\right)\)

\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-12\right)\)

\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)

\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)

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

a) \(x^3-13x-12\)

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

\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-12\right)\)

\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)

\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)

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

b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen

c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha

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

Phương trình thuần nhất đẳng cấp bậc 8 bạn nha :D

c) \(-\frac{x^4}{4}+2x^2y^3-4y^6=-\left(\frac{x^4}{4}-2x^2y^3+4y^6\right)=-\left[\left(\frac{x^2}{2}\right)^2-2.\frac{x^2}{2}.2y^3+\left(2y^3\right)^2\right]=-\left(\frac{x^2}{2}-2y^3\right)\)

a) \(x^4+64\)

= \(\left(x^2\right)^2+8^2\)

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

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

=\(x^2+6x+1x+9+3\)

=\(\left(x^2+6x+9\right)+\left(x+3_{}\right)\)

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

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

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

c) \(x^2-x-6\)

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

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

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

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

a)    \(3x^2-7x+2\)

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

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

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

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

\(=ax^2+a-a^2x-x\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(ax-1\right)\)

(x-căn của 3)(x+căn của 3)

\(x^2-3\)

\(=x^2-\left(\sqrt{3}\right)^2\)

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

/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0

⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0

⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0

Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:

(t-18).t +17=0

⇔t2−18t+17=0⇔t2−18t+17=0

⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0

\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)

\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)

\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)

Đặt ....

a)A=(x²+2x)+9x²+18x+20

=(x²+2x)+9(x²+2x)+20

Đặt t=x²+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x²+2x vào đc:

10(x²+2x+2)

b sai đề