a) Ta có:  x 2 = 2 2  nên x = 2.

b) Ta có: x 2 = 5 2 nên x = 5.

c) Ta có:  3 x 5 = 3  nên  x 5 = 1 . Do đó x = 1.

d) Ta có:  6 x 3 = 48  nên  x 3 = 8 . Do đó x = 2.

e) Ta có:  x - 1 2 = 2 2  nên  x - 1 = 2 . Do đó x = 3.

f) Ta có:  x + 1 2 = 5 2  nên x +1 = 5. Do đó x = 4.

g) Ta có:  x - 1 3 = 3 3  nên  x - 1 = 3 . Do đó x = 4.

h) Ta có:  x + 1 3 = 4 3 nên x +1 = 4. Do đó x = 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.

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

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

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

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

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

a) \(\left(x-1\right)^3\)

\(=x^3-3x^2+3x-1\)

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

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

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

a)219-7(x+1)=100

7(x+1)=219-100

x+1=119/7

x+1=17

x=17-1

x=16

b)(3x-6)*3=64

3x-6=64/3

3x=64/3+6

3x=82/3

x=82/3:3

x=82/9

a) 219-7(x+1)=100

219-7x-7=100

-7x=100-219+7

-7x=-112

x=16

b) (3x-6).3=64

3x-6=64/3

3x=82/3

x=82/9

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4