a.1+3+5+7+...+2x+1=225

[(2x+1-1):2+1]x(2x+1+1):2=225

(x+1)x(2x+2):2=225

(x+1)x(x+1)=225

(x+1)²=225

(x+1)²=15²

x+1=15

x=14

_______________

b. 130-[5.(9-x)+43]=47

    5.(9-x)+43=83

  5.(9-x)=40

9-x=8

x=1

_______________

c.16^x<32⁴

2^4x<2²⁰

=>x∈{0;1;2;3;4}

a: =>(x+1)^2=225

=>x+1=15

=>x=14

b: =>[5*(9-x)+43]=130-47=83

=>5(9-x)=40

=>9-x=8

=>x=1

c: =>2^4x<2^20

=>4x<20

=>0<x<5

a) 2x-13=15

2x       = 15+13

2x      =  18

x        = 18:2

x        =9

c. /3x-7/=2

+) Nếu 3x-7=2 thì  x = 3

+)Nếu 3x-7=-2 thì x = 5/3 không thuộc Z

Vậy x = 3

a ) 16^x : 4^x = 16

     (16:4)^x=16

     4^x=4² hoặc      (-4)²

           Vậy x=2

b) ( 2x + 1 )³ = -64

     ( 2x + 1 )³ = (-4)³

     2x+1=-4

    2x=-5

     x=\(-\frac{5}{2}\)

Vậy x=\(-\frac{5}{2}\)

c ) ( 3x - 1 )² = 16

     ( 3x - 1 )² = 4²=(-4)²

            Suy ra:3x-1=4;3x=5;x=\(\frac{5}{3}\)

                        3x-1=-4;3x=-3;x=-1

Vậy x=-1;\(\frac{5}{3}\)

d ) 9^x+1 = 3²

     9^x+1 = 9¹

     x+1=1

    x=0

Vậy x=0

a) 16^x : 4^x = 16 
=> ( 16 : 4 )^x = 16 
=> 4^x = 16 
<=> 4^x = 4²
=> x = 2 
b) ( 2x + 1 ) ³ = -64 
=> ( 2x + 1 )³ = ( -4 )³
=> 2x + 1 = -4 
=> 2x = -4 - 1 
=> 2x = -5 
=> x = \(-\frac{5}{2}\)
c) ( 3x - 1 )² = 16 
\(\Rightarrow\begin{cases}\left(3x-1\right)^2=4^2\\\left(3x-1\right)^2=\left(-4\right)^2\end{cases}\)\(\Rightarrow\begin{cases}3x-1=4\\3x-1=-4\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}\)
d) 9^x+1 = 3² 
=> (3² ) ^x+1 = 3² 
=> 3^2.(x+1) = 3²
=> 2(x + 1 ) = 2 
=> x + 1 = 1 
=> x = 0 

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 \(x=\dfrac{5}{4}\)

1) 3x = 45 + 15 = 60

x = 60 : 3 = 20

2) 5x = 50 - 35 = 15

x = 15 : 5 = 3

3) (2x - 5) + 17 = 6

2x - 5 = 6 - 17

2x - 5 = -11

2x = -11 + 5

2x = -6

x = -6 : 2 = -3

4) 10 - 2(4 -3x) = -4

2(4 - 3x) = -4 - 10 = -14

4 - 3x = -14 : 2 = -7

3x = -7 - 4 = -18

x = -18 : 3 = -6

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^