tìm x (simple)
1/27=(1/9)^x
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a) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^{22}\)
\(\left(\frac{1}{3}\right)^{2\cdot x}=\left(\frac{1}{3}\right)^{66}\)
\(\Rightarrow2\cdot x=66\)
\(x=66\div2\)
\(x=33\)
Vậy x = 33
b) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}^3\right)\right]^6\)
\(\left(\frac{2}{3}\right)^{2\cdot x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2\cdot x=18\)
\(x=18\div2\)
\(x=9\)
Vậy x = 9
ta có :
\(\frac{x+1}{3}+\frac{x+1}{9}+\frac{x+1}{27}+\frac{x+1}{81}=\left(x+1\right)\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)
\(=\left(x+1\right)\times\frac{40}{81}=\frac{56}{81}\text{ nên }x+1=\frac{56}{40}=\frac{7}{5}\)
vậy \(x=\frac{7}{5}-1=\frac{2}{5}\)
\(x+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}=2\)
\(\Leftrightarrow x+\dfrac{121}{81}=2\)
hay \(x=\dfrac{41}{81}\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-\dfrac{2}{3}\)
=>x*1/3=-8/3
hay x=-8
\(\left(\dfrac{4}{9}\right)^{x+1}=\left(\dfrac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\dfrac{2}{3}\right)^{2x+2}=\left(\dfrac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x+2=18\Leftrightarrow2x=16\Leftrightarrow x=8\)
mũ 3
dễ mà!
\(\frac{1}{27}=\left(\frac{1}{9}\right)^x\)
\(\Rightarrow\frac{1}{27}=\left(\frac{1}{3}\right)^{2x}\)
\(\Rightarrow\left(\frac{1}{3}\right)^3=\left(\frac{1}{3}\right)^{2x}\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)