42 - 2 ( 32 - 2^x +1) = 10
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\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Leftrightarrow32-2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=32-16=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
\(\Leftrightarrow x+1=4\)
\(\Rightarrow x=3\)
\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
=> x = 4
\(42-2.\left(32-2^{x+1}\right)=10\)
\(2.\left(32-2^{x+1}\right)=42-10\)
\(2.\left(32-2^{x+1}\right)=32\)
\(32-2^{x+1}=32:2\)
\(32-2^{x+1}=16\)
\(2^{x+1}=32-16\)
\(2^{x+1}=16\)
\(2^{x+1}=2^4\)
\(x+1=4\)
\(x=4-1\)
\(x=3\)
\(42-2\left(3x-2^{x+1}\right)=10\)
\(2\left(32-2^{x+1}\right)=32\)
\(32-2^{x+1}=16\)
\(2^{x+1}=16\)
\(2^{x+1}=2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
Vậy x = 3
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a, 3x - 5 = 16...............
=> 3x = 16 + 5 = 21..................
=> x = 21 : 3 = 7..........
\(b,42-2\left(32-2^{x+1}\right)=10\)
\(\Rightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Rightarrow32-2^{x+1}=32:2=16\)
\(\Rightarrow2^{x+1}=32-16=16=2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1=3\)
42 - 2 (32-2^x +1)=10
,<=>32=64-2.2^x+2
<=>2.2^x=34
<=>2^x=17
đề sai r