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) 5^x+5^{x + 2} = 650

=> 5^x+ 5^x.5² = 650

=> 5^x(1+ 5²) = 650

=> 5^x.26 = 650

=> 5^x = 650:26

=> 5^x = 25

=> 5^x = 5²

=> x = 2

Vậy 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 = 162

=> x ko có giá trị

Vậy x ko tìm đc giá trị thỏa mãn đề bài.

Ta Có ;

a. 5^x + 5^x+2 = 650 

=> 5^x ( 1 + 25 ) = 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 = 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)

bạn kia làm đúng rồi 

k tui nha 

thank

Bài 1:

a: Ta có: \(48751-\left(10425+y\right)=3828:12\)

\(\Leftrightarrow y+10425=48751-319=48432\)

hay y=38007

b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)

\(\Leftrightarrow2367-y=1222\)

hay y=1145

Bài 2: 

Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

a,\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Leftrightarrow\left(2x-1\right).\left[1-\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\1-\left(2x-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=1\\2x-1=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=2\\2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)

\(b,5^x+5^{x+1}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x.\left(1+5^2\right)\)\(=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=650\div26\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

\(c,3^{x-1}+5.3^{x-1}=162\)

\(\Leftrightarrow3^{x-1}.\left(1+5\right)=162\)

\(\Leftrightarrow3^{x-1}.6=162\)

\(\Leftrightarrow3^{x-1}=162\div6\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow3^{x-1}=3^3\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=3+1\)

\(\Leftrightarrow x=4\)

thanks nhìu

\(a,5^x+5^{x+2}=650\Leftrightarrow5^x+5^x+5^2=650\Leftrightarrow5^x.26=650\Leftrightarrow5^x=5^2\Leftrightarrow x=2\) 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

bn đăng ít thoi

a)5^x+5^x+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)\(3^{x-1}+5\cdot3^{x-1}=162\)

\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)

\(\Rightarrow3^{x-1}\cdot6=162\)

\(\Rightarrow3^{x-1}=27\)

\(\Rightarrow3^{x-1}=3^3\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x=4\)

c và d bài 1 có tại đây Câu hỏi của Vương Hàn - Toán lớp 7 | Học trực tuyến