Tìm x
\(5.3^{3x+2}-27^{x+1}=486\)
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\(3^{x-1}+5.3^{x-1}=486\)
\(\Rightarrow3^{x-1}.\left(5+1\right)=486\)
\(3^{x-1}.6=486\)
\(3^{x-1}=81\)
\(3^{x-1}=3^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
<=> 3x-1 +5.3x-1 =486
<=> 3x-1 . (1+5) = 486
<=> 3x-1 . 6 =486
<=> 3x-1 = 486 :6
<=> 3x-1 = 81
<=> 3x-1 = 34
<=> X-1=4
<=> X = 4+1
<=> X=5
3x-1 + 5 . 3x-1 = 486
3x-1 . ( 5 + 1 ) = 486
3x-1 . 6 = 486
3x-1 = 486 : 6
3x-1 = 81
3x-1 = 34
=> x - 1 = 4
x = 4 + 1
x = 5
Vậy x = 5
3x-1 + 5.3x-1 = 486
1.3x-1 + 5.3x-1 = 486
=> 3x-1.( 1 + 5) = 486
3x-1. 6 = 486
3x - 1 = 486 : 6
3 x - 1 = 81
3 x - 1 = 34
=> x - 1 = 4
x = 4 + 1
x = 5
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\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x\cdot2^2-2^x=96\)
\(\Rightarrow2^x\left(2^2-1\right)=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x\cdot3=96\)
\(\Rightarrow2^x=96:3\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(5^x+5^{x+1}=750\)
\(\Rightarrow5^x+5^x\cdot5=750\)
\(\Rightarrow5^x\left(1+5\right)=750\)
\(\Rightarrow5^x\cdot6=750\)
\(\Rightarrow5^x=750:6\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
\(2^{x+3}+2^x=144\)
\(\Rightarrow2^x\cdot2^3+2^x=144\)
\(\Rightarrow2^x\left(2^3+1\right)=144\)
\(\Rightarrow2^x\cdot9=144\)
\(\Rightarrow2^x=144:9\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
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)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
a, 3x-1(1+5)=162
=>3x-1.6=162
=>3x-1=27
=>3x-1=33
=>x-1=3
=>4
b, x.(3+x)=0
=>x=0 hoặc 3+x=0
=>x=0 hoặc x= -3
Theo đề ta có:
\(5.3^{3x+2}-27^{x+1}=486\)
\(\Rightarrow5.27^x.9-27^x.27=486\)
\(\Rightarrow27^x\left(45-27\right)=486\)
\(\Rightarrow27^x.18=486\)
\(\Rightarrow27^x=\frac{486}{18}\)
\(\Rightarrow27^x=27\)
\(\Rightarrow27^x=27^1\)
\(\Rightarrow x=1\)
vậy \(x=1\)