a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

1:

a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)

b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)

c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)

Bài 2: tất cả đều ở dạng tích rồi mà

\(1,2a^2+2b^2=2\left(a^2+b^2\right)\)

\(2,x^4+13-6x^2+4y+y^2\)

\(=\left(x^4-6x^2+9\right)+\left(y^2+4y+4\right)\)

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

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

c: \(x^6-3x^5+3x^4-x^3=\left(x^2-x\right)^3\)

\(a,=\left(x-4\right)^2\\ b,=\left(\dfrac{1}{2}xy^2+1\right)^2\)

\(\dfrac{x^2+y}{y^2+x}\)-> (x*x+y)/(y*y+x)

\(x^2+\dfrac{y}{y^2}+x\) -> x*x+y/y*y+x 

x*x và y*y có thể thay thế bằng sqr(x) và sqr(y)

a) \(x^2-6x-y^2-4y+5=x^2-6x+9-y^2-4y-4\)

\(=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)=\left(x-3\right)^2-\left(y+2\right)^2\)

b) \(4a^2-12a-b^2+2b+8=4a^2-12a+9-b^2+2b-1\)

\(=\left(4a^2-12a+9\right)-\left(b^2-2b+1\right)=\left(2a-3\right)^2-\left(b-1\right)^2\)

x² - 6x - y² - 4y + 5

= ( x² - 6x + 9 ) - ( y² + 4y + 4 )

= ( x - 3 )² - ( y + 2 )²

4a² - 12a - b² + 2b + 8

= ( 4a² - 12a + 9 ) - ( b² - 2b + 1 )

= ( 2a - 3 )² - ( b - 1 )²

a) 

\(=x^2-6x+9-y^2-4y-4\) 

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

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

b) 

\(=4a^2-12a+9-b^2+2b-1\) 

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

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

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

b) \(y^3-13=\left(y-\sqrt{13}\right)\left(y^2+\sqrt{13}y+13\right)\)

c) \(2x^2-4=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\)

d) \(\left(x-1\right)^3-\left(y+1\right)^3=\left(x-1-y-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]=\left(x-y-2\right)\left(x^2-2x+1+xy-y+x-1+y^2+2y+1\right)=\left(x-y-2\right)\left(x^2+y^2-x+y+xy+1\right)\)

a. x² - 2

<=> x² - \(\left(\sqrt{2}\right)^2\)

<=> (x - \(\sqrt{2}\))(x + \(\sqrt{2}\))

b. y³ - 13 

<=> y³ - \(\left(\sqrt[3]{13}\right)^3\)

<=> \(\left(y-\sqrt[3]{13}\right)\left[y^2+\sqrt[3]{13}y+\left(\sqrt[3]{13}\right)^2\right]\)

c. 2x² - 4

<=> \(\left(x\sqrt{2}\right)^2\) - 2²

<=> \(\left(x\sqrt{2}-2\right)\left(x\sqrt{2}+2\right)\)

d. (x - 1)³ - (y + 1)³

\(\Leftrightarrow\left[\left(x-1\right)-\left(y+1\right)\right]\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]\)

\(\Leftrightarrow\left(x-1-y-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]\)

\(\Leftrightarrow\left(x-y-2\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y+1\right)+\left(y+1\right)^2\right]\)