Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`#3107.\text {DN}`
\(3^{x+2}+4\cdot3^{x+1}+3^{x-1}=6^6\)
`=> 3^x*3^2 + 4*3^x*3 + 3^x * 1/3 = 6^6`
`=>3^x*(3^2 + 12 + 1/3) = 6^6`
`=> 3^x * 64/3 = 6^6`
`=> 3^x = 6^6 \div 64/3`
`=> 3^x = 2187`
`=> 3^x = 3^7`
`=> x = 7`
Vậy, `x = 7.`
Ta có 3^(x+2) +4.3^(x+1) +3^(x-1) =6^6
=> 3^x . 3^2 + 4.3^x.3+ 3^x.1/3 = 46656
=> 3^x( 9+12+1/3) =46656
=> 3^x .64/3 =46656
=> 3^x =2187
=>3^x = 3^7
=> x=7
Vậy x=7
\(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(\Rightarrow3^{x-1}\left(3^3+4.3^2+1\right)=3^6.2^6\)
\(\Rightarrow3^{x-1}.64=3^6.64\)
\(\Rightarrow3^{x-1}=3^6\)
\(\Rightarrow x-1=6\Leftrightarrow x=7\)
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)
\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)
\(\Leftrightarrow x=8\)
b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)
\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)
a. x=8
b. x=13
còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.
Lời giải:
1.
$3^{x+2}+4.3^{x+1}=7.3^6$
$3^{x+1}.3+4.3^{x+1}=7.3^6$
$3^{x+1}(3+4)=7.3^6$
$3^{x+1}.7=7.3^6$
$\Rightarrow 3^{x+1}=3^6$
$\Rightarrow x+1=6$
$\Rightarrow x=5$
2.
$5^{x+4}-3.5^{x+3}=2.5^{11}$
$5^{x+3}.5-3.5^{x+3}=2.5^{11}$
$5^{x+3}(5-3)=2.5^{11}$
$2.5^{x+3}=2.5^{11}$
$\Rightarrow 5^{x+3}=5^{11}$
$\Rightarrow x+3=11$
$\Rightarrow x=8$
3.
$4^{x+3}-3.4^{x+1}=13.4^{11}$
$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$
$4^{x+1}.16-3.4^{x+1}=13.4^{11}$
$13.4^{x+1}=13.4^{11}$
$\Rightarrow 4^{x+1}=4^{11}$
$\Rightarrow x+1=11$
$\Rightarrow x=10$
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
a) (2x - 3)2 = 16
=> (2x - 3)2 = 42
=> \(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
=> \(\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy ...
b) 2.3x + 2 + 4. 3x + 1 = 10.36
=> 2.3x + 1. 3 + 4.3x + 1 = 10 . 36
=> 6.3x + 1 + 4.3x + 1 = 10.36
=> (6 + 4).3x + 1= 10.36
=> 10.3x + 1= 10.36
=> 3x + 1= 36
=> x + 1 = 6
=> x = 6 - 1
=> x = 5
\(a,\left(2x-3\right)^2=16\)
\(\Rightarrow\left(2x-3\right)^2=4^2\)
\(\Rightarrow2x-3=4\)
\(2x=4+3\)
\(2x=7\)
\(x=7:2\)
\(x=\frac{7}{2}\)
\(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(3^{x-1}.3^3+4.3^{x-1}.3^2+3^{x-1}=6^6\)
\(3^{x-1}.\left(27+9.4+1\right)=6^6\)
\(3^{x-1}.\left(27+36+1\right)=2^6.3^6\)
\(3^{x-1}.64=3^{x-1}.2^6=3^6.2^6\)
\(\Rightarrow\)\(3^6=3^{x-1}\Rightarrow x=7\)
3^x.3^2+4.3^x.3+3^x:3=6^6
3^x.9+4.3^x.3+3^x/3=36
nhóm lại mà làm nhé.
k nha