Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

Bài 2:

7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)

9) \(\left(a+b\right)^2-\left(2a-b\right)^2=\left(a+b+2a-b\right)\left(a+b-2a+b\right)=3a\left(-a+2b\right)\)

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

11) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2a\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2a\left(3a^2+b^2\right)\)

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

x³+64=x³+4³=(x+4)(x²-4x+4²)=(x+4)(x²-4x+16)

\(x^3+64=x^3+4^3=(x+4)(x^2-3x+16)\)

\(10x-25-x^2=-\left(x^2-10x+25\right)\)

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

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²

\(2x^2-7x+3\)

\(=2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)\)

Vậy thôi đâu cần dùng HĐT

2x²-7x+3

=2x²-x-6x+3

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

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

\(=\left(4-a+b\right)\left(4+a-b\right)\)

\(\Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)

\(x^2-10x+25=0\)

\(x^2-10x+5^2=0\)

\(\left(x-5\right)^2=0\)

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

\(x^3-6x^2y+12xy^2-8y^3=\left(x-2\right)^3\)

a: \(\left(a-2b\right)^2-4b^2\)

\(=\left(a-2b\right)^2-\left(2b\right)^2\)

=(a-2b-2b)(a-2b+2b)

=a(a-4b)

b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

c: \(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b+2\right)\left(a+b-2\right)\)

d: \(\left(a+3b\right)^2-9b^2\)

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

=(a+3b-3b)(a+3b+3b)=a(a+6b)

e: \(\left(x-3\right)^3-27\)

\(=\left(x-3-3\right)\left\lbrack\left(x-3\right)^2+3\left(x-3\right)+9\right\rbrack\)

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

f: \(\left(x+1\right)^3-125\)

\(=\left(x+1-5\right)\left\lbrack\left(x+1\right)^2+5\left(x+1\right)+25\right\rbrack\)

\(=\left(x-4\right)\left(x^2+2x+1+5x+5+25\right)=\left(x-4\right)\left(x^2+7x+31\right)\)

dấu ^ là mũ nha mn