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a, không có x :v
b, (2x + 1)3 = 729
=> (2x + 1)3 = 93
=> 2x + 1 = 9
=> 2x = 8
=> x = 4
vậy_
c, 97 : x = 97
=> x = 97 : 97
=> x = 1
vậy_
d, x5 : 83 = 64
=> x5 : 83 = 82
=> x5 = 82.83
=> x5 = 85
=> x = 8
vậy_
a) 3x : 3 = 81
3x = 27 = 33
=> x = 3
b) (2x+1)3 = 729 = 93
=> 2x + 1 = 9
2x = 8
x = 4
c) 97: x = 97
x = 1
d) x5 : 83 = 64 = 82
x5 = 83.82 = 85
=> x = 8
3x+3*32x-1*3x=729
\(\Rightarrow3^{x+3+2x-1+x}=3^6\)
\(\Rightarrow3^{4x+2}=3^6\)
\(\Rightarrow4x+2=6\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-4}{3}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
3^ x . 3^ 2= 729
3^ x . 9 = 729
3^ x = 729: 9
3^x =81
vậy x = 3^3
5^ x . 625 = 3125
5^x = 3125:625
5^x = 5
vậy x = 1
( 2x+1 )^ 3 = 27
( 2x+1) ^ 3 = 3^3
vậy 2x+1 = 3
vậy x = 0
\(3^{x+3}.3^{2x-1}.3^x=729\)
\(3^{x+3+2x-1+x}=729\)
\(3^{4x+2}=3^6\)
\(\Rightarrow4x+2=6\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
Tks bạn