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

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

Lời giải:

a. 

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

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

Đặt: \(x^2-6x+1=a;x^2+1=b\)

Khi đó đa thức này có dạng:

\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)

\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)

Thay lại a và b thì được:

\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)

\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)

\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)

Vậy ...

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`