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\(B=\frac{9^{12}\cdot4^8}{27^7\cdot2^{17}}\)
\(B=\frac{\left(3^2\right)^{12}\cdot\left(2^2\right)^8}{\left(3^3\right)^7\cdot2^{17}}\)
\(B=\frac{3^{24}\cdot2^{16}}{3^{21}\cdot2^{17}}\)
\(B=\frac{3^3}{2}\)
\(B=\frac{27}{2}\)
\(B=\frac{9^{12}.4^8}{27^7.2^{17}}\)
\(\Rightarrow B=\frac{\left(3^2\right)^{12}.\left(2^2\right)^8}{\left(3^3\right)^7.2^{17}}\)
\(\Rightarrow B=\frac{3^{24}.2^{16}}{3^{21}.2^{17}}\)
\(\Rightarrow B=\frac{3^3}{2}=\frac{27}{2}\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Vậy \(x\in\left\{\pm7\right\}\)
\(\left(2x-5\right)^2=0,81\)
\(\left(2x-5\right)^2=0,9^2\)
\(\Rightarrow2x-5=0,9\)
\(2x=0,9+5\)
\(2x=5,9\)
\(x=5,9:2\)
\(x=2,95\)
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\(\left(x-\frac{1}{3}\right)^3=0,027\)
\(\left(x-\frac{1}{3}\right)^3=0,3^3\)
\(\Rightarrow x-\frac{1}{3}=0,3\)
\(x=0,3+\frac{1}{3}\)
\(x=\frac{19}{30}\)
\(\left(2x-5\right)^2=0,81\)
\(\Rightarrow2x-5=0,9\)
\(\Rightarrow2x=5,9\)
\(\Rightarrow x=2,95\)
\(\left(x-\frac{1}{3}\right)^3=0,027\)
\(\Rightarrow x-\frac{1}{3}=0,3\)
\(\Rightarrow x=\frac{19}{30}\)
a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(x+\frac{1}{3}=\frac{-2}{3}\)
\(x=-1\)
b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)
\(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\left(\frac{5}{3}\right)^2\)
\(\frac{1}{3}x+\frac{4}{3}=\frac{5}{3}\)
\(\frac{1}{3}x=\frac{1}{3}\)
\(x=1\)
c) \(2^x+2^{x+1}=24\)
\(2^x+2^x.2=24\)
\(2^x.\left(1+2\right)=24\)
\(2^x.3=24\)
\(2^x=8\)
\(2^x=2^3\)
\(x=3\)
a, (x+1/3)^3 = -8/27
=>(x+1/3)^3 = (-2/3)^3
=>x+1/3 = -2/3
=>x = -1
b, (1/3x+4/3)^2 = 25/9
=>(1/3x+4/3)^2 = (5/3)^2
=>(1/3x+4/3) = 5/3
=>1/3x = 1/3
=> x = 1
c, 2^x + 2^x+1 = 24
=>2^x + 2^x . 2 = 24
=>2^x.(1+2) = 24
=>2^x . 3 = 24
=>2^x =8
=>2^x = 2^3
=> x = 3
\(\Rightarrow\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)+\dfrac{1}{2}\left(2x+1\right)=\dfrac{-13}{3}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}+x+\dfrac{1}{2}=\dfrac{-13}{3}\)
\(\Rightarrow\dfrac{4}{3}x=\dfrac{-13}{3}\Rightarrow x=\dfrac{-13}{4}\)
\(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}\)\(\text{và }3x-5y+6z=9\)
MÌNH ĐANG CẦN GẤP GIÚP MÌNH NHA
\(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}\)\(\Leftrightarrow\frac{3\left(x-1\right)}{15}=\frac{5\left(y-2\right)}{15}=\frac{6\left(z-2\right)}{12}\)
\(\Leftrightarrow\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\).Áp dụng tc dãy tỉ số "=" nhau ta có:
\(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}=\frac{\left(3x-3\right)-\left(5y-10\right)+\left(6z-12\right)}{15-15+12}=\frac{9-5}{12}=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}\frac{3x-3}{15}=\frac{1}{3}\Rightarrow x=\frac{8}{3}\\\frac{5y-10}{15}=\frac{1}{3}\Rightarrow y=3\\\frac{6z-12}{12}=\frac{1}{3}\Rightarrow z=\frac{8}{3}\end{cases}}\)
b)\(\frac{27}{3^x}=3\)
\(\Rightarrow3^x=27:3\)
\(\Rightarrow3^x=9\Rightarrow x=2\)
còn câu a mình dg suy nghĩ
câu a hok biết làm
b) 27/3x = 3
=> 27 = 3x x 3
33 = 3x+1
=> 3 = x+1
=> x=2
câu a hình như sai đề phải
Bài làm
a) Ta có: \(\left(-\frac{3}{2}\right)^x=\frac{9}{4}\)
=>\(\left(-\frac{3}{2}\right)^x=\left(-\frac{3}{2}\right)^2\)
Vậy x = 2
b) Ta có: \(\left(-\frac{2}{3}\right)^x=-\frac{8}{27}\)
=> \(\left(-\frac{2}{3}\right)^x=\left(-\frac{2}{3}\right)^3\)
Vậy x = 3
# Học tốt #
=>3^x(1+3^2)=3^34+3^36
=>3^x.10=3^34.10
=>3^x=3^34
=>x=34