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VH
3
2 tháng 7 2019
3-1 . 3x + 5.3x - 1 = 162
=> 3x - 1 + 5.3x - 1 = 162
=> 3x - 1. (1 + 5) = 162
=> 3x - 1 . 6 = 162
=> 3x - 1 = 162
=> 3x - 1 = 27
=> 3x - 1 = 33
=> x - 1 = 3
=> x = 3 + 1
=> x =4
22 tháng 7 2016
\(f\)) \(32^{-x}.16^x=1024\)
\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)
\(\Leftrightarrow2^{4x-5x}=2^{10}\)
\(\Leftrightarrow2^{-x}=2^{10}\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(g\)) \(3^{x-1}.5+3^{x-1}=162\)
\(3^{x-1}.\left(5+1\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)
\(i\)) \(5^x+5^{x+2}=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=650:26\)
\(5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
IA
12 tháng 7 2021
c, \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}=5^{11}\)
\(\Leftrightarrow x+3=11\)
\(\Leftrightarrow x=8\)
Vậy x = 8
d, \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+2^{x+4}+2^{x+5}=480\)
\(\Leftrightarrow2^x\left(1+2+2^2+2^3+2^4+2^5\right)=480\)
\(\Leftrightarrow2^x\cdot63=480\)
\(\Leftrightarrow2^x=\frac{160}{21}\)
\(\Leftrightarrow x\approx2,93\)
20 tháng 8 2020
a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
PN
20 tháng 8 2020
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d
S
15 tháng 7 2019
\(3^x+5\times3^{x-1}=256\)
\(\Rightarrow3^x+5\times3^x\div3=256\)
\(\Rightarrow3^x+5\times3^x\times\frac{1}{3}=256\)
\(\Rightarrow3^x\times\left(1+5+\frac{1}{3}\right)=256\)
\(\Rightarrow3^x\times\frac{19}{3}=256\)
Đến đây mk chịu, đề bài sai hay sao bạn ạ.
~Study well~
#JDW
KD
8
6 tháng 12 2019
81:3^x=3
=> 3^x =81:3
=>3^x=27
=> x=3
3^x+1.2=162
=> 3^x+1=162:2
=> 3^x+1= 81
=> x+1=4
x=3
3^x-1 + 5.3^x-1 = 162
3^x-1. (1+5) = 162
3^x-1 =27
3^x-1 = 3^3
x-1 =3
x = 4
Vậy x=4
\(3^{x-1}+5.3^{x-1}=162\)
\(\Rightarrow3^{x-1}.\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}.6=162\)
\(\Rightarrow3^{x-1}=162:6=27=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1=4\)