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a, Ta có: 3 x = 3 2 nên x = 2
b, Ta có: 5 x = 5 3 nên x = 3
c, Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1
d, Ta có: 6 x - 1 = 6 2 nên x - 1 = 2, đo đó x = 3
e) Ta có: 3 2 x + 1 = 3 3 nên 2x +1 = 3, do đó x = 1
f) Ta có: x 50 = x nên x 50 - x = 0 , do đó x x 49 - 1 = 0 = 0
Vì thế x = 0 hoặc x = 1
a) x = 0
b) x = -1
c) x = -9
d) x = 24
e) x = 2 hoặc x = -4
f) x = 5 hoặc x = -3
`#3107.101107`
a)
\(5\left(x-1\right)^3=40\\\Rightarrow\left(x-1\right)^3=40\div5\\ \Rightarrow\left(x-1\right)^3=8\\ \Rightarrow\left(x-1\right)^3=2^3\\ \Rightarrow x-1=2\\ \Rightarrow x=2+1\\ \Rightarrow x=3\)
Vậy, `x = 3`
b)
\(3^{2x+1}+9^x=324?\\ \Rightarrow3^{2x}\cdot3+3^{2x}=324\\ \Rightarrow3^{2x}\cdot\left(3+1\right)=324\\ \Rightarrow3^{2x}\cdot4=324\\ \Rightarrow3^{2x}=81\\ \Rightarrow3^{2x}=3^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
c)
\(5^x-13=3\cdot2^2\\ \Rightarrow5^x-13=12\\ \Rightarrow5^x=12+13\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
Vậy, `x = 2`
d)
\(8^x+2^{3x+1}=192\\ \Rightarrow2^{3x}+2^{3x}\cdot2=192\\ \Rightarrow2^{3x}\left(1+2\right)=192\\ \Rightarrow2^{3x}\cdot3=192\\ \Rightarrow2^{3x}=64\\ \Rightarrow2^{3x}=2^6\\ \Rightarrow3x=6\\ \Rightarrow x=2\)
Vậy, `x = 2.`
+) \(24+\left(x-6\right)=135\)
\(\Rightarrow x-6=135-24=111\)
\(\Rightarrow x=111+6=117\)
+) \(2019-2\cdot\left(3x-2\right)=19\)
\(\Rightarrow2\cdot\left(3x-2\right)=2019-19=2000\)
\(\Rightarrow3x-2=2000:2=1000\)
\(\Rightarrow3x=1000+2=1002\)
\(\Rightarrow x=1002:3\)
\(\Rightarrow x=334\)
+) \(3^x+3^{x+1}=324\)
\(\Rightarrow3^x+3^x\cdot3=324\)
\(\Rightarrow3^x\cdot\left(1+3\right)=324\)
\(\Rightarrow3^x\cdot4=324\) \(\Rightarrow3^x=324:4=81\)
\(\Rightarrow3^x=3^4\) \(\Rightarrow x=4\)
\(3^x\cdot4=324\)
\(\Leftrightarrow3^x=\dfrac{324}{4}=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(3^x\cdot4=324\)
\(\Rightarrow3^x=324:4\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
a) \(2x^3-32x=0\)
\(2x\left(x^2-16\right)=0\)
\(2x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
b) \(\left(3x-2\right)^2-\left(x+5\right)^2=0\)
\(\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)
\(\left(2x-7\right)\left(4x+3\right)=0\)
\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
c) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x+5\right)\left(2x-5-x-7\right)=0\)
\(\left(2x+5\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)
\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)
\(\Leftrightarrow3^{3x+1}\cdot4=324\)
\(\Leftrightarrow3^{3x+1}=81\)
\(\Leftrightarrow3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(\Rightarrow x=1\)