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3x-1+5.3x-1=162
=>3x-1.(1+5)=162
=>3x-1.6=162=>3x-1=162:6=27=33
=>x-1=3=>x=4
5x+5x+2=650
=>5x+5x.52=650
=>5x.(1+25)=650
=>5x.26=650=>5x=650:26=25=52
=>x=2
a) 5X+5X+2=650
=>5X.1+5X.52=650
=>5X.(1+52)=650
=>5X.26=650
=>5x=650:26
=>5X=25
=>5X=52
=>X=2
b) 3X-1+5.3X-1=162
=>3X-1.1+5.3X-1=162
=>3X-1.(1+5)=162
=>3X-1.6=162
=>3X-1=162:6
=>3X-1=27
=>3X-1=33
=>3X=33+1
=>3X=34
=>X=4
=> \(5^x+5^x.5^2=650\)
\(5^x\left(1+5^2\right)=650\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25;x=2\)
5X+5X+2=650
=>5X.1+5X.52=650
=>5X.(1+52)=650
=>5X.26=650
=>5x=650:26
=>5X=25
=>5X=52
=>X=2
3X-1+5.3X-1=162
=>3X-1.1+5.3X-1=162
=>3X-1.(1+5)=162
=>3X-1.6=162
=>3X-1=162:6
=>3X-1=27
=>3X-1=33
=>3X=33+1
=>3X=34
=>X=4
\(5^x+5^{x+2}=650\Leftrightarrow5^x+5^x.5^2=650\)
\(5^x=650:26=25\Leftrightarrow x=2\)
th1: \(\left(\frac{y}{3}-5\right)^{2008}-\left(\frac{y}{3}-5\right)^{2000}=0\)
\(\left(\frac{y}{3}-5\right)^{2000}.\left[\left(\frac{y}{3}-5\right)^8-1\right]=0\)
\(=>\orbr{\begin{cases}\left(\frac{y}{3}-5\right)^{2000}=0\\\left(\frac{y}{3}-5\right)^{2008}-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}\frac{y}{3}=5=>y=15\\\frac{y}{3}=6=>y=18,\frac{y}{3}=4=>y=12\end{cases}}\)
Vậy ...
P/S: cái đoạn\(\left(\frac{y}{3}-5\right)^{2008}-1=0\)vì số mũ chẵn nên y=18 hay bằng 12 nha!
\(3^{x-1}+5.3^{x-1}=162\)
\(3^{x-1}\left(1+5\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\) là giá trị cần tìm