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

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)
=\(y^3+27t^3\)
 

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

    c) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

2) \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\Rightarrow x^2+6x+9-x^2+4=11\)

\(\Rightarrow6x=-2\Rightarrow x=-\dfrac{1}{3}\)

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

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

Đề bài không chính xác, biểu thức này không viết được dưới dạnh tích

a. y⁴ - 14y² + 49

Gọi y² là t, ta có:

t² - 14t + 49

<=> t² - 14t + 7²

<=> (t - 7)²

Thay x² = t

<=> (x² - 7)²

b. \(\dfrac{1}{4}-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)

\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

c. x⁴ - 16

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

d. x² - 9

<=> x² - 3²

<=> (x - 3)(x + 3)

Bài 1: 

a: \(y^2-14y^2+49=\left(y-7\right)^2\)

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

c: \(x^4-16=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)

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

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

a)-x^3+3x^2-3x+1

=-(x³-3x²+3x-1)

=-(x-1)³

b)8-12x+6x^2-x^3

=2³-3.2².x+3.2.x²-x³

=(2-x)³