Lời giải:

a. $21-2(x+6)=15$

$2(x+6)=21-15=6$

$x+6=6:2=3$

$x=3-6=-3$

b.

$3^{x+1}=27=3^3$

$\Rightarrow x+1=3$

$\Rightarrow x=2$

c.

$14-x=36$

$x=14-36=-22$

d.

$-3x+25=-2$

$-3x=-2-25=-27$

$x=(-27):(-3)=9$

e.

$37-2(x+6)=29$

$2(x+6)=37-29=8$

$x+6=8:2=4$

$x=4-6=-2$

h.

$2^{x-1}=16=2^4$

$x-1=4$

$x=4+1=5$

Th1: 2x+3 ≥ 0
Khi đó: |2x+3| =x+2
 (2x+3)= x+2
- 2x+3= x+2
-2x-x= 2-3
 x= -1
Th2: 2x+3 < 0
Khi đó: |2x+3|=x+2
 -(2x+3) = x +2
 -2x-3 = x+2
 -3x = 5
 x=-5/3

Vậy x= -1

      x= -5/3

Lớp 6 cugx học dạng v nè

Câu b nha 

a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.

b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.

c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.

d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.

a) \(2^x\cdot4=16\)

\(\Rightarrow2^x=16:4\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

b) \(3^x\cdot3=243\)

\(\Rightarrow3^x=243:3\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4

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

a: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

=>\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

=>9x+7=16

=>9x=9

=>x=1

b: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=17\)

=>\(x^3-25x-x^3-8=17\)

=>-25x=8+17=25

=>x=-1

c: \(x^3-12x^2+48x-64=0\)

=>\(x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3=0\)

=>\(\left(x-4\right)^3=0\)

=>x-4=0

=>x=4

\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!

20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)