`a)(x-6)^2-(x+6)^2=12`

`<=>(x-6-x-6)(x-6+x+6)=12`

`<=>-12.2x=12`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2`

`b)36x^2-12x+1=81`

`<=>(6x-1)^2=81`

`<=>(6x-1-9)(6x-1+9)=0`

`<=>(6x-10)(6x+8)=0`

`<=>(3x-5)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x-6)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2-6x+x-6=0`

`<=>x(x-6)+x-6=0`

`<=>(x-6)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

a) \(=\left(6x\right)^2-2.6x.1+1=\left(6x-1\right)^2\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(=\left(3x-y\right)^2-25=\left(3x-y-5\right)\left(3x-y+5\right)\)

d) \(=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

a,x(x+y)-5x-5y

=x(x+y)-5(x+y)

=(x+y)(x-5)

b,3x-5y-6ax+10ay

=(3x-6ax)-(5y-10ay)

=3x(1-2a)-5y(1-2a)

=(1-2a)(3x-5y)

c,a²-6a-b²+6b

=(a²-b²)-(6a-6b)

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

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

d,100a²-20a-2b-b²

=(100a²-b²)-(20a+2b)

=(10a-b)(10a+b)-2(10a+b)

=(10a+b)(10a-b-2)

e,36x²-12x+1-b²

=(36x²-12x+1)-b²

=(6x-1)²-b²

=(6x-1-b)(6x-1+b)

f,x²-z²+y²-2xy

=(x²-2xy+y²)-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

Bài 2:

b: \(=\left(x^2+x+4+3x\right)\left(x^2+x+4+5x\right)\)

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

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

c: \(=25x^2-49y^2-\left(5x+7y\right)\)

=(5x+7y)(5x-7y-1)

d: \(8x^3-36x^2+54x-27=\left(2x-3\right)^3\)

⇔127-(2x+9)³=2

⇔(2x+9)³=125

⇔(2x+9)³=5³

⇒2x+9=5

⇔2x=-4

⇔x=-2 

Vậy x=-2

ii) Biến đổi vế trái (VT), ta có:

\(36\times\dfrac{2}{3}=\dfrac{72}{3}=24\\ 280\times\dfrac{3}{5}=\dfrac{840}{5}=168\)

\(\dfrac{36\times2}{3}=24;\dfrac{280\times3}{5}=168\)

36×2/3=72/3=24

280×3/5=840/5=168

nhá