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\(\dfrac{2}{3}\cdot3^{x+1}-7\cdot3^x=-405\)
\(\Rightarrow3^x\cdot\left(\dfrac{2}{3}\cdot3-7\right)=-405\)
\(\Rightarrow3^x\cdot\left(2-7\right)=-405\)
\(\Rightarrow3^x\cdot-5=-405\)
\(\Rightarrow3^x=-405:-5\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy: \(x=4\)
\(\Leftrightarrow3\cdot\dfrac{2}{3}\cdot3^x-7\cdot3^x=-405\)
=>\(-5\cdot3^x=-405\)
=>3^x=81
=>x=4
`@` `\text {Ans}`
`\downarrow`
`a)`
`210 \div x - 1/2 = 20,5`
`=> 210 \div x = 20,5 + 1/2`
`=> 210 \div x =21`
`=> x = 210 \div 21`
`=> x = 10`
Vậy, `x = 10.`
`b)`
`7 * 3^x + 20*3^x = 3^25`
`=> 3^x * (7+20) = 3^25`
`=> 3^x * 27 = 3^25`
`=> 3^x * 3^3 = 3^25`
`=> 3^x = 3^25 \div 3^3`
`=> 3^x = 3^22`
`=> x = 22`
Vậy, `x = 22.`
a) \(210:x-\dfrac{1}{2}=20,5\)
\(\Rightarrow210:x=20,5+\dfrac{1}{2}\)
\(\Rightarrow210:x=21\)
\(\Rightarrow x=\dfrac{210}{21}\)
\(\Rightarrow x=10\)
b) \(7\cdot3^x+20\cdot3^x=3^{25}\)
\(\Rightarrow3^x\cdot\left(7+20\right)=3^{25}\)
\(\Rightarrow3^x\cdot27=3^{25}\)
\(\Rightarrow3^x\cdot3^3=3^{25}\)
\(\Rightarrow3^{x+3}=3^{25}\)
\(\Rightarrow x+3=25\)
\(\Rightarrow x=25-3\)
\(\Rightarrow x=22\)
a,\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(x=-\dfrac{13}{15}\)
b,\(\left(3x+1\right)^3=-27\)
\(3x+1=-3\)
\(3x=-4\)
\(x=-\dfrac{4}{3}\)
a ) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\\dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{13}{10}\\\dfrac{3}{2}x=\dfrac{17}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)
Vậy .......
b ) \(\left(3x+1\right)^3=-27\)
\(\Leftrightarrow3x+1=-3\)
\(\Leftrightarrow3x=-4\Leftrightarrow x=-\dfrac{4}{3}\)
Vậy ..........
c ) \(7.3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow7.3^{x-1}-3^{x-1}.3^3=-540\)
\(\Leftrightarrow3^{x-1}\left(7-27\right)=-540\)
\(\Leftrightarrow-20.3^{x-1}=-540\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow x=4\).
Vậy ..........
\(7\cdot3^{x-1}-3^{x+2}=540\)
=>\(7\cdot3^x\cdot\dfrac{1}{3}-3^x\cdot9=540\)
=>\(3^x\left(\dfrac{7}{3}-9\right)=540\)
=>\(3^x\cdot\dfrac{-20}{3}=540\)
=>\(3^x=-540:\dfrac{20}{3}=-540\cdot\dfrac{3}{20}=-27\cdot3=-81\)
=>\(x=log_3\left(-81\right)\)
7.3x-1 - 3x+2 = -540
<=> 7.3x : 3 - 3x.32 = -540
<=> 3x.(7 : 3 - 32) = -540
<=> 3x = 81
<=> 3x = 34
<=> x = 4
Ta có :
\(7.3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow\)\(7.3^x:3-3^x.3^2=-540\)
\(\Leftrightarrow\)\(\frac{7}{3}.3^x-9.3^x=-540\)
\(\Leftrightarrow\)\(3^x\left(\frac{7}{3}-9\right)=-540\)
\(\Leftrightarrow\)\(3^x.\frac{-20}{3}=-540\)
\(\Leftrightarrow\)\(3^x=\left(-540\right):\frac{-20}{3}\)
\(\Leftrightarrow\)\(3^x=81\)
\(\Leftrightarrow\)\(3^x=3^4\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=4\)
\(7\times3^{x-1}-3^{x+2}=-540\)
\(7\times3^{x-1}-3^{x-1}\times3^3=-540\)
\(3^{x-1}\left(7-3^3\right)=-540\)
\(3^{x-1}\left(7-27\right)=-540\)
\(3^{x-1}\times\left(-20\right)=-540\)
\(3^{x-1}=\left(-540\right)\div\left(-20\right)\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(x=4\)
Chúc bn học tốt
\(7\cdot3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow3^x\left[7\cdot\left(-3\right)-3^2\right]=-540\)
\(\Leftrightarrow3^x\cdot\left(-30\right)=-540\)
\(\Leftrightarrow3^x=18\)
Ta có : \(7.3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow\left(7-3^3\right).3^{x-1}=-540\)
\(\Leftrightarrow\left(7-27\right).3^{x-1}=-540\)
\(\Leftrightarrow-20.3^{x-1}=-540\)
\(\Leftrightarrow3^{x-1}=\left(-540\right):\left(-20\right)\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
Vậy \(x=4\)