3^x-1+3^x+3^x+1=39 tìm x
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39.(5 - 2\(x\)) = 39
5 - 2\(x\) = 39 : 39
5 - 2\(x\) = 1
2\(x\) = 5 - 1
2\(x\) = 4
\(x=4:2\)
\(x=2\)
Vậy \(x=2\)
\(c,\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\\ \Rightarrow\left(x-2\right)\left(1-x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ d,\Rightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\\ \Rightarrow\left(x^2+3+1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2+4=0\left(vô.nghiệm\right)\\x+1=0\end{matrix}\right.\Rightarrow x=-1\)
a. ( x - 1/2 ) x 5/3= 7/4 - 1/2
(x -1/2 ) x 5/3 = 5/4
x - 1/2 = 5/4 : 5/3
x - 1/2 = 3/4
x = 3/4 + 1/2
x = 5/4
b . 6/13 : ( 1/2 + x ) = 15/39
1/2 + x = 6/13- 15/39
1/2 + x = 1/13
x = 1/13 - 1/2
x = -11/26
c. 16/3 : (1/2+ x ) = 15/39
1/2 + x = 16/3 - 15/69
1/2 + x = 353/69
x = 353/69 - 1/2
x = 637/138
=> 3x - 1 (3 + 1 + 32) = 39
=> 3x - 1.13 = 39
=> 3x - 1 = 39/13
=> 3x - 1 = 3
=> x - 1 = 1
=> x = 1 + 1
=> x = 2
3x+3x+1+3x+2=243.39
3x+3x.3+3x.9=9477
3x.(1+3+9)=9477
3x.13=9477
3x=729=36
x=6
3x + 3x+1 + 3x+2 = 243.39
<=> 3x + 3x.3 + 3x.32 = 243.39
<=> 3x( 1 + 3 + 32 ) = 243.39
<=> 3x.13 = 243.39
<=> 3x = 243.3 = 729
<=> 3x = 36
<=> x = 6
a) \(5+3^{x+1}=86\)
\(=>3^{x+1}=86-5\)
\(=>3^{x+1}=81=3^4\)
\(=>x+1=4\) ( cùng cơ số )
\(=>x=4-1\)
\(=>x=3\)
b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(=>15:\left(x+2\right)=\left(27+3\right):10\)
\(=>15:\left(x+2\right)=30:10=3\)
\(=>x+2=15:3\)
\(=>x+2=5\)
\(=>x=5-2\)
\(=>x=3\)
c) \(\left(9x+2\right).4=80\)
\(=>9x+2=80:4\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:9\)
\(=>x=2\)
d) \(\left(245-x\right)+7^2=14\)
\(=>\left(245-x\right)+14=14\)
\(=>245-x=14-14\)
\(=>245-x=0\)
\(=>x=245-0\)
\(=>x=245\)
3x-1 + 3x + 3x+1 = 39
3x . \(\dfrac{1}{3}\) + 3x + 3x . 3 = 39
3x . ( \(\dfrac{1}{3}\) + 1 + 3 ) = 39
3x . \(\dfrac{13}{3}\) = 39
3x = 39 : \(\dfrac{13}{3}\)
3x = 9
3x = 32
x = 2
Vậy x = 2
3^x là j v