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Bài 1:
a) \(4^{x+2}+4^x=68\)
\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)
\(\Rightarrow4^x\cdot17=68\)
\(\Rightarrow4^x=\dfrac{68}{17}\)
\(\Rightarrow4^x=4\)
\(\Rightarrow4^x=4^1\)
\(\Rightarrow x=1\)
b) \(5\cdot2^{x+4}-3\cdot2^x=308\)
\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)
\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)
\(\Rightarrow2^x\cdot77=308\)
\(\Rightarrow2^x=\dfrac{308}{77}\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
c) \(4\cdot3^{x+1}+7\cdot3^x=513\)
\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)
\(\Rightarrow3^x\cdot19=513\)
\(\Rightarrow3^x=\dfrac{513}{19}\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
d) \(5^{x+4}-5^x=3120\)
\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)
\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)
\(\Rightarrow5^x\cdot624=3120\)
\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)
\(\Rightarrow5^x=5\)
\(\Rightarrow5^x=5^1\)
\(\Rightarrow x=1\)
f) \(3\cdot4^{2x+1}-16^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)
\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)
\(\Rightarrow4^{2x}\cdot11=2816\)
\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)
\(\Rightarrow4^{2x}=256\)
\(\Rightarrow\left(2^2\right)^{2x}=2^8\)
\(\Rightarrow2^{4x}=2^8\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Bài 2:
\(2^x+124=5^y\)
\(\Rightarrow5^y-2^x=124\)
\(\Rightarrow5^y-2^x=125-1\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)
Vậy: ....
\(7.3^x+20.3^x=3^{25}\Rightarrow3^x\left(7+20\right)=3^{25}\)
\(\Rightarrow3^x.3^3=3^{25}\Rightarrow3^x=3^{22}\Rightarrow x=22\)
\(7\cdot3^x+20\cdot3^x=3^{25}\)
\(3^x\cdot\left(20+7\right)=3^{25}\)
\(3^x\cdot27=3^{25}\)
\(3^x\cdot3^3=3^{25}\)
\(3^x=3^{25}:3^3\)
\(3^x=3^{22}\)
\(x=22\)
Đề có phải như thế này không vậy bạn?
\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\)
\(3\cdot3^{x+1}+4\cdot3^{x+1}=7\cdot3^6\)
\(\left(3+4\right)\cdot3^{x+1}=7\cdot3^6\)
\(7\cdot3^{x+1}=7\cdot3^6\)
x + 1 = 6
x = 6 - 1 = 5
Vậy x = 5
a, \(4^7.3^4.9^6:6^{13}\)
\(=\left(2^{14}.3^4.3^{12}\right):\left(2^{13}.3^{13}\right)\)
\(=2^{14}:2^{13}.3^{16}:3^{13}\)
\(=2.3^3=54\)
b, \(2^3.3^2-5^{16}:25^7\)
\(=72-5^{16}:5^{14}\)
\(=72-5^2=47\)
4^7.3^4.9^6:6^13=4^7.3^4.(3^2)^6:6^13
=4^7.3^16:3^13.2^13
=(2^2)^7.3^16:3^13.2^13
=2^14.3^16:3^13.2^13
=2.3^3
=54
Câu 1:
a) \(-\dfrac{3}{7}-\left(\dfrac{2}{3}-\dfrac{3}{7}\right)=\dfrac{-3}{7}-\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{-2}{3}\)
Câu 2:
b) \(\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{4}{5}-\dfrac{1}{3}\cdot\dfrac{6}{5}\right)=\dfrac{2}{15}:\left[\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\right]=\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{-2}{5}\right)=\dfrac{2}{15}:\dfrac{-2}{15}=\dfrac{2}{-2}=-1\)
5.3x = 5.34
=> x = 4
5.34 = 7.35 - 2.35
= 35 . (7 - 2)
= 5. 35
=> 3x = 35
=> x = 5
`(25+x)+3.4=7.3^2`
`(25+x)+12=63`
`25+x=63-12=51`
`x=51-25=26`