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a) \(5^x+5^{x+2}=650\)
\(5^x+5^x.5^2=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=25\)
\(5^x=5^2\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(3^{x-1}+5.3^{x-1}=162\)
\(3^{x-1}.\left(1+5\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(x=3+1\)
\(x=4\)
Vậy x = 4
a) 5x+5x + 2 = 650
=> 5x+ 5x.52 = 650
=> 5x(1+ 52) = 650
=> 5x.26 = 650
=> 5x = 650:26
=> 5x = 25
=> 5x = 52
=> x = 2
Vậy x = 2
b) 3x-1 + 5.3x-1 = 162
=> 3x-1(1+5) = 162
=> 3x-1. 6 = 162
=> 3x-1 = 162
=> x ko có giá trị
Vậy x ko tìm đc giá trị thỏa mãn đề bài.
a)\(5^x+5^{x+2}=650\Rightarrow5^x+5^2.5^x=650\Rightarrow5^x+25.5^x=650\Rightarrow26.5^x=650\)\(5^x=25\Rightarrow5^x=5^2\Rightarrow x=2\)
b) \(3^{x-1}+5.3^{x-1}=162\Rightarrow6.3^{x-1}=162\Rightarrow3^{x-1}=27=3^3\)x-1=3 nên x=4
a ) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
b ) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}\left(1+5\right)=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\Leftrightarrow x=4\)
Ta Có ;
a. 5x + 5x+2 = 650
=> 5x ( 1 + 25 ) = 650
=> 5x . 26 = 650
=> 5x = 25
=> x = 2
b. 3x-1 + 5.3x-1 =162
=> 3x-1 ( 1 + 5 ) = 162
=> 3x-1 . 6 = 162
=> 3x-1 = 27
=> x - 1 = 3
=> x = 3+1 = 4
CHO TÍCH NHA !
Tìm x,biết:
a) (2x-4)4= 81 b) (x-1)5 = -32 c) (2x-1)6=(2x-1)
a,\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right).\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\1-\left(2x-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=1\\2x-1=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=2\\2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)
\(b,5^x+5^{x+1}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x.\left(1+5^2\right)\)\(=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=650\div26\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
\(c,3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}.\left(1+5\right)=162\)
\(\Leftrightarrow3^{x-1}.6=162\)
\(\Leftrightarrow3^{x-1}=162\div6\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=3+1\)
\(\Leftrightarrow x=4\)
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
=> \(5^x+5^x.5^2=650\)
\(5^x\left(1+5^2\right)=650\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25;x=2\)
a: (x-3)2=49
=>x-3=7 hoặc x-3=-7
=>x=10 hoặc x=-4
b: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
=>x=0 hoặc x=1
c: \(\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
3x-1+5.3x-1=162
=>3x-1.6=162
=>3x-1=162:6
=>3x-1=27=33
=>x-1=3
=>x=4
a)(2x-1)6=(2x-1)8
<=>(2x-1)8-(2x-1)6=0
<=>(2x-1)6[(2x-1)2-1)]=0
TH1.2x-1=0=>2x=1=>x=1/2
TH2.(2x-1)2-1=0=>(2x-1)2=1=>2x-1=1=>2x=2=>x=1
b)5x+5x+2=650=>5x(1+52)=650=>5x=25=>x=2
c)3x-1+5.3x-1=162=>3x-1(1+5)=162=>3x-1=27=>3x=9=32=>x=2
a/ ( 2x - 1 ) 6 = ( 2x - 1 ) 8
Mà chỉ có 0 và 1 là thỏa mãn trên
Nên 2x - 1 = 0 thì 2x = 1 (sai)
2x - 1 = 1 thì 2x = 2 => x = 1 (t/m)
a) 5^x + 5^x+2 = 650
<=> 5^x. (1+5^2)= 650
<=> 5^x. 26= 650
<=> 5^x= 25
=> x= 2
b) 3^x-1 + 5.3^x-1 = 162
<=> 3^x-1. (1+5)= 162
<=> 3^x+1. 6= 162
<=> 3^x+1= 27
=> x+1= 3
=> x= 2