mình giải ý c 

5(x+27)=200

(x+27)=200:5

(x+27)=40

x=40-27

x=13

1/ ( x-3) ²=16

\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

2/ (3x-1)³=8

\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)

3/ (x-11)³=-27

\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)

phần 4 mình ko rõ đề

4) \(x^3-3x^2+3x-1=-64\)

\(\Rightarrow x^3-3x^2+3x+63=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(6x^2+18x\right)+\left(21x+63\right)=0\\ \Rightarrow x^2\left(x+3\right)+6x\left(x+3\right)+21\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x^2+6x+21\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x^2+6x+21=0\end{matrix}\right.\)

\(x+3=0\\ \Rightarrow x=-3\)

\(x^2+6x+21=0\\ \Rightarrow\left(x^2+6x+9\right)+12=0\\ \Rightarrow\left(x+3\right)^2+12=0\)

Vì \(\left(x+3\right)^2\ge0;12>0\Rightarrow\left(x+3\right)^2+12>0\Rightarrow x^2+6x+21vônghiệm\)

Vậy \(x=-3\)

a) 2.3^x=162

=> 3^x=162:2

=> 3^x=81

=> 3^x=3⁴

=> x=4

b) 3^x+3^x+2=810

=> 3^x+3^x.3²=810

=> 3^x.(3²+1)=810

=> 3^x=810:10

=> 3^x=81

=> 3^x=3⁴

=> x=4

2.3^x=162

  3^x= 162:2

  3^x= 81

 3^x= 3^4

=>  x=4

2.3^x=162

3^x=162:2

3^x=81

x=4

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3^2.6.3^x+1=162

9.6.3^x+1=162

  6.3^x+1=162:9

  6.3^x+1=18

     3^x+1=18:6

     3^x+1=3

     3^x+1=3^1

Suy ra x=0

vậy x=o thỏa mãn đề bài

a)x- 72 =8

x= 8+72

x=80

b) 2.3^x=162

3x=162:2

3x=81

x=81:3

x=27

c)

3^x+3^x+2=810

=> 3^x.(1+3²)=810

=> 3^x.(1+9)=810

=> 3^x.10=810

=> 3^x=810:10

=> 3^x=81

=> 3^x=3⁴

=> x=4

a, X-72=8

<=>X=8+72

<=>X=80

b,2.3^x=162

<=>3^x=162:2

<=>3^x=81

<=>3^x=3⁴ =>X=4

c,3^x+3^x+2=810

<=>3^x.1+3^x.3²=810

<=>3^x(1+3²)=810

<=>3^x+10=810

<=>3^x=810:10=81

Mà 3⁴=81 hoặc (-3)⁴=81

 vậy x=4 hoặc x=-4 .

Mk sẽ giải từng câu 

\(a)\) \(\left(3x+1\right)\left(x-2\right)>0\)

Trường hợp 1 : 

\(\hept{\begin{cases}3x+1>0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}3x>-1\\x>2\end{cases}\Leftrightarrow}\hept{\begin{cases}x>\frac{-1}{3}\\x>2\end{cases}}}\)

\(\Rightarrow\)\(x>2\)

Trường hợp 2 : 

\(\hept{\begin{cases}3x+1< 0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}3x< -1\\x< 2\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{-1}{3}\\x< 2\end{cases}}}\)

\(\Rightarrow\)\(x< \frac{-1}{3}\)

Vậy \(x>2\) hoặc \(x< \frac{-1}{3}\) thì \(\left(3x+1\right)\left(x-2\right)>0\)

Chúc bạn học tốt ~ 

a) (3x+1).(x-2)>0

TH1: 3x+1>0                  TH2: x-2>0

3x > -1                                    x>2

x>-1/3

Vậy x>2

\(b)\) \(\left(2x-1\right)\left(x+2\right)\le0\)

Trường hợp 1 : 

\(\hept{\begin{cases}2x-1\le0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}2x\le1\\x\ge-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le\frac{1}{2}\\x\ge-2\end{cases}}}\)

\(\Rightarrow\)\(-2\le x\le\frac{1}{2}\)

Trường hợp 2 : 

\(\hept{\begin{cases}2x-1\ge0\\x+2\le0\end{cases}\Leftrightarrow\hept{\begin{cases}2x\ge1\\x\le-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{1}{2}\\x\le-2\end{cases}}}\)

\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)

Vậy \(-2\le x\le\frac{1}{2}\) thì \(\left(2x-1\right)\left(x+2\right)\le0\)

Chúc bạn học tốt ~ 

162:3^4=2

vậy x = 4

\(\dfrac{162}{3^x}=2\)

\(\Rightarrow3^x=\dfrac{162}{2}=81\)

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

\(\Rightarrow x=4\)

162/3^x=2

=>3^x=162:2

=>3^x=81

=>3^x=3^4

=>X=4

vậy x=4