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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.
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)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
a) (2x-1)3 = 27
(2x-1)3 = 93
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2
b) (2x-1)4 = 81
(2x-1)4 = (\(\pm\)34)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé
a,\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right).\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\1-\left(2x-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=1\\2x-1=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=2\\2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)
\(b,5^x+5^{x+1}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x.\left(1+5^2\right)\)\(=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=650\div26\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
\(c,3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}.\left(1+5\right)=162\)
\(\Leftrightarrow3^{x-1}.6=162\)
\(\Leftrightarrow3^{x-1}=162\div6\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=3+1\)
\(\Leftrightarrow x=4\)
\(a,5^x+5^{x+2}=650\Leftrightarrow5^x+5^x+5^2=650\Leftrightarrow5^x.26=650\Leftrightarrow5^x=5^2\Leftrightarrow x=2\) x=2
b,Với x=0 khi đó 3^0-1+5.3^0-1=2 (loại)
Với x=1 khi đó 3^1+5.3^1=18 (loại)
Với x=2 khi đó 5.3^x-1>16 (loại)
Vậy không có x thỏa mãn
a, \(\left(x+1\right)^2=169\)
\(\left(x+1\right)^2=13^2\)
\(x+1=13\)
\(x=13-1\)
\(x=12\)
1.
a) \(\left(x+1\right)^2=169\)
⇒ \(x+1=\pm13\)
⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{12;-14\right\}.\)
b) \(\left(x+3\right)^3=-\frac{1}{27}\)
⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)
⇒ \(x+3=-\frac{1}{3}\)
⇒ \(x=\left(-\frac{1}{3}\right)-3\)
⇒ \(x=-\frac{10}{3}\)
Vậy \(x=-\frac{10}{3}.\)
c) \(\left(2x-4\right)^4=\frac{1}{625}\)
⇒ \(2x-4=\pm\frac{1}{5}\)
⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)
Còn câu d) bạn làm tương tự như mấy câu trên.
Chúc bạn học tốt!
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)\(5^x+5^{x+2}=650\)(=)\(5x.\left(1+25\right)=650\)(=)\(5^x.26=650\)(=)\(5^x=25\)=>x=2
b)\(3^{x-1}+5.3^{x-1}=162\)(=)\(3^{x-1}.\left(1+5\right)=162\)(=)\(3^{x-1}.6=162\)(=)\(3^{x-1}=27\)(=)\(3^{x-1}=3^3\)=>x-1=3(=)x=2
c)\(4^x+4^{x+3}=4160\)(=)\(4^x.\left(1+64\right)=4160\)(=)\(4^x.65=4160\)(=)\(4^x=64\)(=)\(4^x=4^3\)
=>x=3
học tốt