tìm x biết 8.3^x+3.2^x-6^x=24
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\(10+2\cdot x=4^5:4^3\)
\(\Rightarrow10+2\cdot x=4^{5-3}\)
\(\Rightarrow10+2\cdot x=4^2\)
\(\Rightarrow10+2\cdot x=16\)
\(\Rightarrow2\cdot x=16-10\)
\(\Rightarrow2\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
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\(2\cdot x-6^2:18=3\cdot2^2\)
\(\Rightarrow2\cdot x-36:18=12\)
\(\Rightarrow2\cdot x-2=12\)
\(\Rightarrow2\cdot x=12+2\)
\(\Rightarrow2\cdot x=14\)
\(\Rightarrow x=\dfrac{14}{2}=7\)
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\(70-5\cdot\left(x-3\right)=3\cdot2\)
\(\Rightarrow70-5\cdot\left(x-3\right)=6\)
\(\Rightarrow70-5x+15=6\)
\(\Rightarrow-5x+15=6-70\)
\(\Rightarrow-5x+15=64\)
\(\Rightarrow-5x=64-15\)
\(\Rightarrow-5x=49\)
\(\Rightarrow x=-\dfrac{49}{5}\)
a) 120 - 5 . ( x + 2 ) = 45
5 . (x + 2) = 120 - 45
5 . (x + 2) = 75
x + 2 = 75 : 5
x + 2 = 15
x = 17
b) ( 2.x - 3 )2 = 49
( 2.x - 3 )2 = 72
( 2.x - 3 ) = 7
2x = 10
x = 5
\(\Leftrightarrow\dfrac{3\cdot2^{x+1}}{2^{x+3}}=40\\ \Leftrightarrow\dfrac{3}{2^2}=40\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
Giang ho dai ca viet nham nhé:
\(5.3^x=8.3^9+7.27^3\)
<=> \(5.3^x=8.3^9+7.\left(3^3\right)^3\) <=> \(5.3^x=8.3^9+7.3^9\)
<=> \(5.3^x=15.3^9\) <=> \(3^x=3.3^9=3^{10}\) => x = 10
nhầm chỉnh lại :
\(5.3^x=8.3^9+7.27^3\Rightarrow5.3^x=8.3^9+7.3^9=15.3^9\Rightarrow15.3^{x-1}=15.3^9\Rightarrow3^{x-1}=3^9\Rightarrow x-1=9\Rightarrow x=10\)
1/5.2^x+1/3.2^x.2=1/5.2^7+1/3.2^7.2
2x(1/5+1/3.2)=2^7(1/5+1/3.2)
=>2^x=2^7
=> x=7
\(8.3^x+3.2^x-6^x=24\)
\(\Rightarrow8.3^x+3.2^x-6^x-24=0\)
\(\Rightarrow\left(8.3^x-2^x.3^x\right)-\left(24-3.2^x\right)=0\)
\(\Rightarrow3^x\left(8-2^x\right)-3\left(8-2^x\right)=0\Rightarrow\left(8-2^x\right)\left(3^x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2^x=8\\3^x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)