a: ĐKXĐ: \(3x^2+6x\ne0\)

=>\(x^2+2x\ne0\)

=>\(x\cdot\left(x+2\right)\ne0\)

=>\(x\notin\left\{0;-2\right\}\)

b: ĐKXĐ: \(x^3+64\ne0\)

=>\(x^3\ne-64\)

=>\(x\ne-4\)

c: ĐKXĐ: \(x^2-1\ne0\)

=>\(x^2\ne1\)

=>\(x\notin\left\{1;-1\right\}\)

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

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

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

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

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

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

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

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

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)

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

\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)

\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)

\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)

mà 7>0

nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)

x³ + 12x² + 48x + 64 = 8x³ - 12x² + 6x - 1

\(\Leftrightarrow\) x³ + 12x² + 48x + 64 - 8x³ + 12x² - 6x + 1 = 0

\(\Leftrightarrow\) -7x³ + 24x² + 42x + 65 = 0

Bn cho đề thế này ai mà giải được :vvv

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => 4x² - 12x + 9 - x² - 10x - 25 = 0 

=> 3x² - 22x - 16 = 0

=> (x - 8)(3x + 2) = 0

=> x - 8 = 0 => x = 8

hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3

Vậy x = 8 ; x = -2/3

c/ => x³ - x² - 4x² + 8x - 4 = 0 

=> x³ - 5x² + 8x - 4 = 0 

=> (x - 2)² (x - 1) = 0

=> (x - 2)² = 0 => x - 2 = 0 => x = 2

hoặc x - 1 = 0 => x = 1 

Vậy x = 2 ; x = 1

1. 179 - ( 55 + 2x ) = 64

             ( 55 +2x)  = 179 - 64

              55 + 2x = 115

                      2x = 115 - 55

                      2x = 60

                        x = 30

a) 8x + 2x = 25 . 2²

   8x + 2x = 25 . 4

   8x + 2x = 100

x ( 8 + 2 ) = 100

x . 10       = 100

     x        = 100 : 10

    x         = 10

b) Theo đề bài ta có :

35 chia hết cho x

=> x thuộc Ư ( 35 )

Ư ( 35 ) = { 1 ; 5 ; 7 ; 35 }

=> x = 1 ; 5 ; 7 ; 35

c) 4^x = 64

Mà 64 = 4³

=> 4^x = 4³

=> x = 3

a) 8x + 2x = 25 . 2² 

(8 + 2) x = 25 . 4

10x = 100

=> x = 100 : 10 = 10

b) 35 chia hết cho x

=> x = {1; 5; 7; 35}

c) 4^x = 64

4^x = 4³

=> x = 3