tìm x
|\(\frac{1}{3}\)x| . |- 2.7| = |9|
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\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
<=>\(\left(\frac{2}{3}\right)^{\frac{n}{2}}=\left(\frac{2}{3}\right)^5\)
<=>\(\frac{n}{2}=5\)
<=>n=10
\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow\left(\frac{2}{3}\right)^{2n}=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow2n=5\Rightarrow n=\frac{5}{2}\)
Vậy n = 5/2
\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)
\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)
\(2^x\cdot\frac{9}{2}=288\)
\(2^x=64\)
\(2^x=2^6\)
\(x=6\)
\(9^x:3^x=3^7\)
\(3^{2x}:3^x=3^7\)
\(3^x=3^7\)
\(x=7\)
\(7^{x+2}+2\cdot7^{x-1}=345\)
\(7^x\cdot7^2+2\cdot7^x:7=345\)
\(7^x\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\cdot\frac{345}{7}=345\)
\(7^x=7\)
\(x=1\)
a) 1/2.2^x + 2^x+2 = 256 + 32
1/2.2^x + 2^2.2^x=288
2^x(1/2+4)= 288
2^x.4,5=288
2^x= 288:4,5
2^x=64=2^6
x=6
\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2
\(9^x:3^x=3^7\)
\(\Rightarrow9:3^x=3^7\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
\(\left|\frac{1}{3}x\right|\cdot\left|-2,7\right|=\left|9\right|\)
=> \(\left|\frac{1}{3}x\right|\cdot\frac{27}{10}=9\)
=> \(\left|\frac{1}{3}x\right|=\frac{10}{3}\)
=> \(\orbr{\begin{cases}\frac{1}{3}x=\frac{10}{3}\\\frac{1}{3}x=-\frac{10}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=10\\x=-10\end{cases}}\)