K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
21 tháng 7 2019
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(5^{x+3}\left(5-3\right)=2.5^{11}\)
\(5^{x+3}.2=2.5^{11}\)
\(5^{x+3}=5^{11}\)
\(x+3=11\)
\(x=8\)
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)
\(4^{x+1}.13=13.4^{11}\)
\(4^{x+1}=4^{11}\)
\(x+1=11\)
\(x=10\)
15 tháng 9 2017
\(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)
\(\Rightarrow4^{x+1}\cdot\left(4^2-3\right)=13\cdot4^{11}\)
\(\Rightarrow4^{x+1}\cdot13=13\cdot4^{11}\)
\(\Rightarrow4^{x+1}=4^{11}\)
\(\Rightarrow x+1=11\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
27 tháng 5 2022
Câu 9:
\(3^x=9^{-6}\cdot27^{-6}\cdot81^8\)
\(\Leftrightarrow3^x=\dfrac{1}{9^6\cdot27^6}\cdot81^8=\dfrac{1}{3^{12}\cdot3^{18}}\cdot3^{32}=3^2\)
=>x=2
Câu 18:
\(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)
\(\Leftrightarrow4^x\cdot64-3\cdot4^x\cdot4=13\cdot4^{11}\)
\(\Leftrightarrow4^x\left(64-3\cdot4\right)=13\cdot4^{11}\)
\(\Leftrightarrow4^x=13\cdot4^{10}\cdot4:52=4^{10}\)
=>x=10
12 tháng 8 2016
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
\(16.4^{x+1}-3.4^{x+1}=13.4^{11}\)
\(\left(16-3\right).4^{x+1}=13.4^{11}\)
\(13.4^{x+1}=13.4^{11}\)
\(\Rightarrow x+1=11\)
\(x=10\)
12 tháng 8 2016
\(4^{x+3}-3.4^{x+1}=13.4^{11}\)
=> \(4^{x+1}.\left(4^2-3\right)=13.4^{11}\)
=> \(4^{x+1}.\left(16-3\right)=13.4^{11}\)
=> \(4^{x+1}.13=13.4^{11}\)
=> \(x+1=11\)
=> \(x=11-1=10\)
28 tháng 9 2016
d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)
<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)
<=> x = -2010
Sửa đề:\(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)
\(\Leftrightarrow4^x\cdot64-3\cdot4^x\cdot4=13\cdot4^{11}\)
\(\Leftrightarrow4^x\cdot52=52\cdot4^{10}\)
\(\Leftrightarrow4^x=4^{10}\)
hay x=10