a)

\(3^x+3^{x+1}+3^{x+2}=117\\ \Leftrightarrow3^x+3.3^x+9.3^x=117\\ 13.3^x=117\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)

b)

 \(3+4\left(x-10\right)=3^2+6\\ \Leftrightarrow3+4\left(x-10\right)=15\\ \Leftrightarrow4\left(x-10\right)=12\\ \Leftrightarrow x-10=3\\ \Leftrightarrow x=13\)

a) \(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x+3^x.3+3^x.3^2=117\)

\(3^x.\left(1+3+3^2\right)=117\)

\(3^x.13=117\)

\(3^x=9\)

\(x=2\)

b) \(3+4\left(x-10\right)=3^2+6\)

\(3+4x-40=9+6\)

\(4x=15+40-3\)

\(4x=52\)

\(x=13\)

a) \(x-2=-6\)

\(x=-6+2\)

\(x=-4\)

b) \(15-\left(x-7\right)=-21\)

\(x-7=36\)

\(x=43\)

c) \(4.\left(3x-4\right)-2=18\)

\(4\left(3x-4\right)=20\)

\(3x-4=5\)

\(3x=9\)

\(x=3\)

d) \(\left(3x-6\right)+3=32\)

\(3x-6=29\)

\(3x=29+6\)

\(3x=35\)

\(x=\frac{35}{3}\)

e) \(\left(3x-6\right).3=32\)

\(3x-6=\frac{32}{3}\)

\(3x=\frac{32}{3}+6\)

\(3x=\frac{50}{3}\)

\(x=\frac{50}{9}\)

f) \(\left(3x-6\right):3=32\)

\(3x-6=96\)

\(3x=102\)

\(x=34\)

g) \(\left(3x-6\right)-3=32\)

\(3x-6=35\)

\(3x=41\)

\(x=\frac{41}{3}\)

h) \(\left(3x-2^4\right).7^3=2.7^4\)

\(\left(3x-2^4\right)=2.7=14\)

\(\left(3x-16\right)=14\)

\(3x=14+16=30\)

\(x=10\)

i) \(\left|x\right|=\left|-7\right|\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

k) \(\left|x+1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

l) \(\left|x-2\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

m) \(x+\left|-2\right|=0\)

\(x+2=0\)

\(x=-2\)

o) \(72-3\left|x+1\right|=9\)

\(3\left|x-1\right|=63\)

\(\left|x-1\right|=21\)

\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)

p) Ta có: \(\left|x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)

mà \(x+1< 0\)

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

\(\Rightarrow x=-2\)

q) \(\left(x-2\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

hok tốt!!

Đề là j vậy ạ???

Thỏ Ruby đề ở trên đấy bạn

Đề là tìm x hay là phân tích đa thức ,....

3x-1 - 2 = 32 + [52 - 3(22 - 1)]

3x-1-2=9+[25-3(4-1)]

3x-1-2=9+(25-3.3)

3x-1-2=9+(25-9)

3x-1-2=9+16

3x-1-2=25

3x-1=25+2

3x-1=27

3x-1=33

=>x-1=3

x=3+1

x=4

2x-1 + 3 = 52

2x-1+3=25

2x-1=25-3

2x-1=22

2.(1 + 3 + 3² + ... + 3ˣ) + 1 = 81

2.(3ˣ⁺¹ - 1)/2 + 1 = 81

3ˣ⁺¹ - 1 + 1 = 81

3ˣ⁺¹ = 81 

3ˣ⁺¹ = 3⁴

x + 1 = 4

x = 4 - 1

x = 3

\(3x-32>-5x+1\)

\(3x+5x>32+1\)

\(8x>33\)

\(x>\frac{33}{8}\)

\(3x-32>-5x+1\)

\(3x>-5x+1+32\)

\(3x>-5x+33\)

\(3x+5x>33\)

\(8x>33\)

=> \(x>\frac{33}{8}\)

Vậy \(x>\frac{33}{8}\)

3x+5x>1+32

8x>33

x>33/8

Hỏi gì đóa?

\(3x-32>-5x+1\)

\(\Leftrightarrow8x>33\Leftrightarrow x>\frac{33}{8}\)

Vậy \(x>\frac{33}{8}\)

\(\left(3x-1\right)^5=32\\ \Rightarrow\left(3x-1\right)^5=2^5\\ \Rightarrow3x-1=2\\ \Rightarrow x=1\)

(3x - 1)^5 = 2^5

3x -1 =2

3x= 2+1

3x =3

x =1

`(3x - 1)^5 = 32`

`<=> (3x - 1)^5 = 2^5`

`<=> 3x - 1 = 2`

`<=> 3x = 3`

`<=> x = 1`

x-32+15-3x=-4x+1-32

x-3x+4x=1-32+32-15

2x=1+32-32-15

2x=1+(32-32)-15

2x=1-15

2x=-14

x=-14:2

x= - 7