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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
g)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{32}{8}=4\)
\(+)\)\(\dfrac{x}{3}=4\Rightarrow x=4\times3=12\)
\(+)\)\(\dfrac{y}{5}=4\Rightarrow y=4\times5=20\)
Vậy \(x=12;y=20\)
h)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\left|-\dfrac{1}{5}\right|\)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{1}{2}-\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{10}-\dfrac{2}{10}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{10}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{3}{10}\) hoặc \(x-\dfrac{1}{3}=\dfrac{-3}{10}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{3}{10}\\x-\dfrac{1}{3}=\dfrac{-3}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}+\dfrac{1}{3}\\x=\dfrac{-3}{10}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{30}+\dfrac{10}{30}\\x=\dfrac{-9}{30}+\dfrac{10}{30}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{30}\\x=\dfrac{1}{30}\end{matrix}\right.\)
\(\dfrac{x}{4}=-\dfrac{5}{3}\)\(\Rightarrow x.3=-5.4\Rightarrow3x=-20\Rightarrow x=\dfrac{-20}{3}\)
\(\dfrac{1}{2}+x=1,5.\dfrac{4}{3}\)\(\Rightarrow\dfrac{1}{2}+x=2\Rightarrow x=2-\dfrac{1}{2}\Rightarrow x=\dfrac{3}{2}\)
\(\dfrac{2}{3}-\dfrac{1}{3}.x=\dfrac{1}{5}\Rightarrow\)\(\dfrac{1}{3}.x=\dfrac{2}{3}-\dfrac{1}{5}\Rightarrow\dfrac{1}{3}.x=\dfrac{7}{15}\Rightarrow x=\dfrac{7}{15}:\dfrac{1}{3}\)\(\Rightarrow x=\dfrac{7}{5}\)
\(\left(x-5,2\right).\dfrac{1}{3}=15\Rightarrow x-5,2=15:\dfrac{1}{3}\Rightarrow x-5,2=45\)\(\Rightarrow x=45+5,2\Rightarrow x=50,2\)
P/s Nhớ tick cho mình nha. Thanks bạn
b: =>2|3x+1|=18
=>|3x+1|=9
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=9\\3x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a. \(x-\left(1,5-7\right)=0,35\\ \Rightarrow x+5,5=0,35\\ \Rightarrow x=-5,14\)
b. \(\left(x-1\right)^5=32\\ \Rightarrow\left(x-1\right)^5=2^5\\ \Rightarrow x-1=2\\ \Rightarrow x=3\)
\(\left(x-1\right)^5+4^3=32\)
\(\Leftrightarrow\left(x-1\right)^5+64=32\)
\(\Rightarrow\left(x-1\right)^5=-32\)
\(\Rightarrow x-1=-2\)
\(\Rightarrow x=-1\)
a)\(\frac{-11}{12}.x+0,25=5\)
\(\Rightarrow-\frac{11}{12}.x=5-0,25=\frac{19}{4}\)
\(\Rightarrow-\frac{11}{12}.x=\frac{19}{4}\)
\(\Rightarrow x=\frac{-57}{11}\)
b)\(\left(x-1\right)^5=-32=-2^5\)
\(\Rightarrow\left(x-1\right)=-2\)
\(\Rightarrow x=-2+1=-1\)
\(\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=-2^5\)
\(x-1=-2\)
\(x=-1\)
Là sao vậy bạn ??