a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)

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

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

a) \(8x^3-\dfrac{1}{125}y^3\)

\(=\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\)

\(=\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\)

\(=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{24}y^2\right)\)

b) \(-x^3+6x^2y-12xy^2+8y^3\)

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

\(=-\left(x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\right)\)

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

a.

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

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

b.

\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)

\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)=\left(x-2\right)\left(x^2+x-8x-8\right)\)

\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)

c.

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)

\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

2.
a) 4x(x-1)-6x+6
= 4x(x-1)-6(x-1)
= (4x-6)(x-1)
3.
a) 6x²-24x=0
    6x(x-4)=0
TH1: 6x=0         TH2: x-4=0
           x=0                     x=4
Vậy x\(\in\){0;4}

2. a. \(4x\left(x-1\right)-6x+6\)

\(=4x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(4x-6\right)\left(x-1\right)\)

3. a. \(6x^2-24x=0\)

\(\Leftrightarrow6x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Bài 3: 

a: \(\Leftrightarrow6x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a: Ta có: \(x^2-6x+9-y^2\)

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

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

b: Ta có: \(x^3+4x^2+4x\)

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

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

c: Ta có: \(4xy-4x^2-y^2+9\)

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

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

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

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

a) \(3x-1=\left(\sqrt{3x}\right)^2-1^2=\left(\sqrt{3x}-1\right)\left(\sqrt{3x}+1\right)\)

b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\)

b) \(4x-25=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4=\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\)

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)

a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)

b: \(ax^2+a-a^2x-x\)

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

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

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

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

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

\(=\left(a-b\right)\left(2a-4b\right)\)

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

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

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