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\(1,x.34-x.10=240\)
\(\Rightarrow x.\left(34-10\right)=240\)
\(\Rightarrow x.24=240\)
\(\Rightarrow x=240:24=10\)
\(2,4^{x+1}=16\)
\(\Rightarrow4^{x+1}=4^2\)
\(\Rightarrow x+1=2\Rightarrow x=1\)
\(3,\left(x+1\right)^3=27=3^3\)
\(\Rightarrow x+1=3\Rightarrow x=2\)
12-(2-x)2=3
(2-x)2=12-3
(2-x)2=9
(2-x)2=32
2-x=3
x=2-3
x=-1
\(\orbr{\frac{\begin{cases}x^4-16=0=>x^4=2^4=>x=+-2\\3^{2x-1}-27=0=>2x-1=3=>x=2\end{cases}}{ }}\)
DS x={-2,2}
a) \(\left(x+5\right)^2=100\Leftrightarrow\orbr{\begin{cases}\left(x+5\right)^2=10^2\\\left(x+5\right)^2=\left(-10\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}x+5=10\\x+5=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-15\end{cases}}}\)
b) \(\left(2x-4\right)^2=0\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
c) \(\left(x-1\right)^3=27\Leftrightarrow\left(x-1\right)^3=3^3\Leftrightarrow x-1=3\Leftrightarrow x=4\)
a) \(\left(x+5\right)^2=100\)
=> \(\orbr{\begin{cases}\left(x+5\right)^2=10^2\\\left(x+5\right)^2=\left(-10\right)^2\end{cases}}\)
=> \(\orbr{\begin{cases}x+5=10\\x+5=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x=5\\x=-15\end{cases}}\)
b) \(\left(2x-4\right)^2=0\)
=> \(2x-4=0\)
=> \(2x=4\)
=> \(x=2\)
c) \(\left(x-1\right)^3=27\)
=> \(\left(x-1\right)^3=3^3\)
=> \(x-1=3\)
=> \(x=4\)
\(4^x=64\)
\(\Rightarrow x=3\)
\(15^x=225\)
\(\Rightarrow x=2\)
\(3^x:9=27\)
\(\Leftrightarrow3^x=243\)
\(\Leftrightarrow x=5\)
\(x^{2018}=0\)
\(\Leftrightarrow x=0\)
\(x^{50}=x\)
\(\Rightarrow x\in\left\{0;1\right\}\)
\(300^x=1\)
\(\Rightarrow x=0\)
`\(12^x=144\)
\(\Rightarrow x=2\)
4x=64=43=> x=3
15x=225=152=> x=2
3x :9 = 27
3x=33x32=35=> x=5
X2018=0=> x=0
X50=x=> x=1 hoặc 0
300x=1=> x=0
12x=144=122=> x=2
x4-1=27
=>x4-1=33
=>x3=33(vì 4-1=3)
=> x=3
3x-1=27
=>3x-1=33
=>x-1=3
=>x=2
a) Ta có: x4-1 = 27
Mà 33 = 27
x3 = 33
Vậy x = 3
b) Ta có: 3x-1 = 27
Mà 33 = 27
=> 3x-1 = 33
=> x - 1 = 3
=> x = 3+1=4
Vậy x = 4