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- a.(3x)2=1/243x33=1/9
3x=1/3 hoặc 3x=-1/3 ( vế 2 ko có x thỏa mãn)
suy ra x=3-1
b.(5x+1)=\(\sqrt{\frac{36}{49}}\)\(\Rightarrow\)5x+1=\(\frac{4}{7}\)hoặc 5x+1=\(\frac{-4}{7}\) | |
\(\Rightarrow\)x=\(\frac{-3}{35}\)hoặc x=\(\frac{-11}{35}\) | |
c.\(\frac{6}{4}\)-10x = \(\frac{4}{5}\)-3x chuyển vế :\(\frac{6}{4}\)-\(\frac{4}{5}\)= -3x + 10x \(\frac{7}{10}\)=7x \(\Rightarrow\)x =\(\frac{7}{10}\):7 \(\Rightarrow\)x= \(\frac{1}{10}\) |
a/ \(3^{x+1}=9^x=3^{2x}\Rightarrow x+1=2x\Leftrightarrow x=1\)
b/ \(2^{3x+2}=4^{x+5}=2^{2x+10}\Rightarrow3x+2=2x+10\Leftrightarrow x=8\)
c/ \(3^{2x-1}=243=3^5\Rightarrow2x-1=5\Leftrightarrow x=3\)
a) \(2^{3x+2}=4^{x+5}\)
\(2^{3x+2}=2^{2x+10}\)
\(\Rightarrow3x+2=2x+10\)
\(3x-2x=10-2\)
\(x=8\)
Vậy x = 8
b) \(3^{2x-1}=243\)
\(3^{2x-1}=3^5\)
\(\Rightarrow2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)
Vậy x = 3
=))
a) \(3^{x-1}=\frac{1}{243}\)
\(3^{x-1}=\left(\frac{1}{3}\right)^5\)
\(\Rightarrow3^{x-1}=3^{-5}\)
\(\Rightarrow x-1=-5\)
\(x=-4\)
b) \(2^x+2^{x+3}=114\)
\(2^x.\left(1+2^3\right)=114\)
\(2^x.9=114\)
\(2^x=\frac{38}{3}\)si đề
\(a,2^x=16\Rightarrow2^x=4^2\Rightarrow x=4\)
\(b,3^{x+1}=9^x\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)
\(c,2^{3x+2}=4^{x+5}\Rightarrow2^{3x+2}=2^{2\left(x+5\right)}\Rightarrow3x+2=2x+10\Rightarrow x=8\)
\(d,3^{2x-1}=243\Rightarrow3^{2x+1}=3^5\Rightarrow2x+1=5\Rightarrow x=2\)
a, 2x = 16
2x = 24
=> x = 4
b, 23x + 2 = 4x + 5
23x + 2 = (22)x + 5
23x + 2 = 22x + 10
=> 3x + 2 = 2x + 10
=> 3x - 2x = 10 - 2
=> x = 8
c, 3x + 1 = 9x
3x + 1 = (32)x
3x + 1 = 32x
=> x + 1 = 2x
=> x = 1
d, 32x -1 = 243
32x - 1 = 35
=> 2x - 1 = 5
2x = 5 + 1
2x = 6 <=> x = 3
Bài giải
\(3^x\cdot3^{x+2}=243\)
\(3^{x+x+2}=243\)
\(3^{2x+1}=243\)
\(3^{2x+1}=3^5\)
\(\Rightarrow\text{ }2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4\text{ : }2\)
\(x=2\)
3^x.3^x+2=243
Suy ra3^x.3^x.3^2=243
3^x.(3^2.1)=243
3^x.9=243
3^x=243:9
3^x=27
Suy ra x=3