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\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ =1-\dfrac{1}{2^{99}}< 1\)
Vậy \(B< 1\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\dfrac{1}{2^{99}}\)
\(\rightarrow B< 1\rightarrowđpcm\)
a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}}\)hoặc \(\hept{\begin{cases}x< -1\\x>2\end{cases}\left(Loai\right)}\)
\(\Leftrightarrow-1< x< 2\)
b) \(\left(x-2\right)\left(x+\frac{1}{2}\right)>0\)
\(\Leftrightarrow\hept{\begin{cases}x-2>0\\x+\frac{1}{2}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2< 0\\x+\frac{1}{2}< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>2\\x>\frac{-1}{2}\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2\\x< \frac{-1}{2}\end{cases}}\)
\(\Leftrightarrow x>2\)hoặc \(x< \frac{-1}{2}\)
Vậy \(\orbr{\begin{cases}x>2\\x< \frac{-1}{2}\end{cases}}\)
a, \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\text{ }\left(x+1\right)\text{ và }\left(x-2\right)\text{ trái dấu}\)
Mà \(x+1>x-2\)
\(\Rightarrow\text{ }\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\) \(\Rightarrow\text{ }\hept{\begin{cases}x>-1\\x< 2\end{cases}}\) \(\Rightarrow\text{ }-1< x< 2\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }1\right\}\)
b, \(\left(x-2\right)\left(x+\frac{1}{2}\right)>0\)
\(\Rightarrow\hept{\begin{cases}x-2>0\\x+\frac{1}{2}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-2< 0\\x+\frac{1}{2}< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x>2\\x>-\frac{1}{2}\end{cases}}\) hoặc \(\hept{\begin{cases}x< 2\\x< -\frac{1}{2}\end{cases}}\)
\(x>2\) hoặc \(x< -\frac{1}{2}\)
x+\(\frac{1}{3}\)=\(\frac{2}{5}\)- (\(\frac{-1}{3}\))
x + \(\frac{1}{3}\)= \(\frac{2}{5}\)+\(\frac{1}{3}\)
x +1/3 =11/15
x= 11/15 -1/3
x= 2/5
b, 5/7-x=1/4 -(-3/5)
5/7 - x = 1/4 +3/5
5/7 - x =17/20
x = 5/7 -17/ 20
x= -19/140
\(\Leftrightarrow x\cdot\dfrac{1}{5}=\dfrac{1}{3}\)
hay \(x=\dfrac{5}{3}\)
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coi bộ khó rùi nha!
a hỏi ông goolge là ra