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a) 2^x . 16^2 = 1024 b) 64 . 4^x = 16^8 c) 2^x = 16
=> 2^x . 256 = 1024 => 64 . 4^x = (4^2) ^ 8 => 2^x = 2^4
=> 2^x = 1024 : 256 => 4^3 . 4^x = 4^16 => x = 4
=> 2^x = 4 => 4^x = 4^16 : 4^3
=> 2^x = 2^2 => 4^x = 4^13
=> x = 13
=> x = 2
a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)
b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)
c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
x - (17 - x) = x - 7
x = x - 7 + 17 - x
x = (-7 + 17) + (x - x)
x = 10 + 0
x = 10
Vậy x = 10.
\(\frac{x+22}{11}+\frac{x+23}{12}=\frac{x+24}{13}+\frac{x+25}{14}\)
\(\Leftrightarrow\left(\frac{x+22}{11}+1\right)+\left(\frac{x+23}{12}+1\right)=\left(\frac{x+24}{13}+1\right)+\left(\frac{x+25}{14}+1\right)\)
\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}=\frac{x+37}{13}+\frac{x+39}{14}\)
\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}-\frac{x+37}{13}-\frac{x+39}{14}=0\)
\(\Leftrightarrow\frac{2184\cdot\left(x+33\right)+2002\cdot\left(x+35\right)-1848\cdot\left(x+37\right)-1716\cdot\left(x+39\right)}{24024}=0\)
\(\Leftrightarrow2184x+72072+2002x+70070-1848x-68376-1716x-66924=0\)
\(\Leftrightarrow622x+6842=0\)
\(\Leftrightarrow x=-11\)
a) Ta có: 5 .(x+4) = 123 - 38
5 .(x+4) = 85
x+4 = 85 : 5
x+4 = 17
x = 17 - 4
x = 13
Câu b: Tương tự
a) 5 . ( x + 4 ) = 123 - 38
5 . ( x + 4 ) = 85
x + 4 = 85 : 5
x + 4 = 17
x = 17 - 4
x = 13
b ) \(3.x-16=2.7^4:7^3\)
3 . x - 16 = 2 . 7
3 . x - 16 = 14
3 . x =14 + 16
3 . x = 30
x = 30 : 3
x = 10
k mik nhé các bn I LVE YOU VERY MUCH