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1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
\(2^{2x+1}=32\)
\(\Leftrightarrow2^{2x+1}=2^5\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
\(a,|2x-2019|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2019=1\\2x-2019=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=2020\\2x=2018\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1010\\x=1009\end{cases}}\)
Vậy ............
\(b,\left(2-x\right)^5=-32\)
\(\Leftrightarrow\left(2-x\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow2-x=-2\)
\(\Leftrightarrow x=4\)
Vậy ..........
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\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
1) 41 - 2x + 1 = 9
-2x + 1 = 9 - 41
-2x + 1 = -32
2x + 1 = 32
2x + 1 = 25
x + 1 = 5
x = 5 - 1
x = 4
mình nhầm :
2x + 1 = 5
=> 2x = 5- 1
=> 2x = 4
=> x = 4 : 2
=> x = 2
Phần trên là đúng nhé!!!!!!
32 = 2^5
=> 2^2x+1=31
=> x = 2