a) (2x-1)³ = 27
(2x-1)³ = 9³
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2

b) (2x-1)⁴ = 81
(2x-1)⁴ = (\(\pm\)3⁴)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé

À phần b) bạn sửa dòng (2x-1)⁴ = (\(\pm\)3⁴) thành (2x-1)⁴ = (\(\pm\)3)⁴ nhé
Mình vừa viết nhầm

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!!

\(2^x.2^1=32:2=16\)

\(2^x.2^1=2^4\)

\(2^x=2^{4-1}=2^3\Rightarrow x=3\)

\(7^x.7^{12}=56-7=49\)

\(7^x.7^{12}=7^2\)

\(7^x=7^{2-12}=7^{-10}\Rightarrow x=-10\)

\(3^x.3^1=162:6=27\)

\(3^x.3^1=3^3\)

\(3^x=3^{3-1}=3^2\Rightarrow x=2\)

3\(x\) - 6 = 35 : 32

3\(x\) - 6 = \(\dfrac{35}{32}\)

3\(x\)      = \(\dfrac{35}{32}\) + 6

3\(x\)      = \(\dfrac{227}{32}\)

  \(x\)      = \(\dfrac{227}{32}\) : 3

  \(x\)    = \(\dfrac{227}{96}\)

tìm x biết 

x⋮4, 15 < x < 35 
Vậy x=?

   @   Linh Phạm:

\(x\) ⋮ 4  và 15 < \(x\) < 35

\(x\) \(⋮\) 4 ⇒ \(x\) \(\in\) B(4) = {0; 4; 8; 12; 16; 20; 24; 28; 32; 36;...;}

mà 15 < \(x\) < 35 nên \(x\) \(\in\) {16; 20; 24; 28; 32}

Vậy: \(x\) \(\in\) {16; 20; 24; 28; 32}

Ta có:\(\frac{12-3x}{32}=\frac{6}{4-x}\)

\(\Leftrightarrow\left(12-3x\right)\left(4-x\right)=32.6\)

\(\Leftrightarrow3\left(4-x\right)\left(4-x\right)=192\)

\(\Leftrightarrow\left(4-x\right)^2=64\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-4\end{cases}}\)

Vậy x = 12 hoặc x = -4

Linz

\(\frac{12-3x}{32}=\frac{6}{4-x}\)

\(\Leftrightarrow\left(12-3x\right)\left(4-x\right)=192\)

\(\Leftrightarrow3\left(4-x\right)\left(4-x\right)=192\)

\(\Leftrightarrow\left(4-x\right)^2=64\)

\(\Rightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}}\)

\(\frac{12-3x}{32}=\frac{6}{4-x}\left(đkxđ:x\ne4\right)\)

\(< =>\left(12-3x\right)\left(4-x\right)=32.6\)

\(< =>3\left(4-x\right)\left(4-x\right)=32.6\)

\(< =>\left(4-x\right)^2=\frac{32.6}{3}=32.2=64\)

\(< =>\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}< =>\orbr{\begin{cases}x=-4\\x=12\end{cases}}}\)

(3x+1)^3=-216=(-6)^3

=>3x+1=-6

=>x=...

3^x-1+5.3^x-1=162

=>3^x-1.(1+5)=162

=>3^x-1.6=162

=>3^x-1=162:6=27

=>3^x-1=3^3

=>x-1=3

=>x=4