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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Vậy \(( - 32{x^5} + 1):( - 2x + 1) = 16{x^4} + 8{x^3} + 4{x^2} + 2x + 1\).
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\Leftrightarrow\frac{2^x}{2}+5.\frac{2^x}{2^2}=\frac{7}{32}\Leftrightarrow2^x\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\Leftrightarrow2^x=\frac{1}{8}=2^{-3}\)
<=> x=-3
1) \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4.49=196\\y^2=4.16=64\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=14,y=8\\x=-14,y=-8\end{matrix}\right.\) (vì \(\dfrac{x}{7}=\dfrac{y}{4}\) nên \(x,y\) cùng dấu)
2) \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)
\(\Leftrightarrow2^{x-1}+\dfrac{5}{2}.2^{x-1}=\dfrac{7}{32}\)
\(\Leftrightarrow2^{x-1}=\dfrac{1}{16}=2^{-4}\)
\(\Leftrightarrow x-1=-4\)
\(\Leftrightarrow x=-3\)
3) \(\left|x+5\right|+\left(3y-4\right)^{2016}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
b: =>2|3x+1|=18
=>|3x+1|=9
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=9\\3x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Ta có :
2x+1 - 2x = 32
=> 2x . ( 2 - 1 ) = 32
=> 2x = 25
=> x = 5