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\(9^{x+1}+5.3^{2x}=324\)
\(9^x.9+5.\left(3^2\right)^x=324\)
\(9^x.9+5.9^x=324\)
\(9^x.\left(5+9\right)=324\)
\(9^x.14=324\)
\(9^x=\frac{324}{14}\)
\(\Rightarrow x\in\varnothing\)
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) \(5^x+5^{x+2}=650\)
\(5^x+5^x.5^2=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=25\)
\(5^x=5^2\)
\(\Rightarrow x=2\)
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.
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)
9x+1-5.32x=324
=>9x.9-(32)x.5=324
=>9x.9-9x.5=324
=>9x(9-5)=324
=>9x.4=324
=>9x=324:4
=>9x=81
=>9x=92
=>x=2
vậy x=2
\(3^{2x}+3^{2x+1}=324\)
\(\Rightarrow3^{2x}+3^{2x}.3=324\)
\(\Rightarrow3^{2x}\left(1+3\right)=324\)
\(\Rightarrow3^{2x}.4=324\)
\(\Rightarrow3^{2x}=324:4=81=3^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2=2\)
32x + 32x + 1 = 324
=> 32x + 32x.3 = 324
=> 32x.(1 + 3) = 324
=> 32x.4 = 324
=> 32x = 324 : 4
=> 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
\(9^{n+1}-5\cdot3^{2n}=324\)
\(9^n\cdot9-5\cdot9^n=324\)
\(9^n\cdot\left(9-5\right)=324\)
\(9^n\cdot4=324\)
\(9^n=324:4=81\)
\(9^n=9^2\)
\(n=2\)
\(9^{x+1}-5.3^{2x}=324=>9^x.9-5.\left(3^2\right)^x=324=>\left(3^2\right)^x.9-5.\left(3^2\right)^x=324\)
\(=>3^{2x}.\left(9-5\right)=324=>3^{2x}=\frac{324}{4}=81=3^4=>2x=4=>x=2\)
vậy x=2
tick nhé