tìm x 32 . 6 . 3x+1=162
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a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
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
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
\(x\) = \(\dfrac{227}{96}\)
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
\(\frac{12-3x}{32}=\frac{6}{4-x}\)
\(\Leftrightarrow\left(12-3x\right)\left(4-x\right)=192\)
\(\Leftrightarrow3\left(4-x\right)\left(4-x\right)=192\)
\(\Leftrightarrow\left(4-x\right)^2=64\)
\(\Rightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}}\)
(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
32 . 6 . 3x+1 = 162
9 . 6 . 3x+1 = 162
54 . 3x+1 = 162
3x+1 = 162 : 54
3x+1 = 3
3x+1 = 31
30+1 = 31
x = 0
Vậy x = 0