2x3^x+1-3^x-1=3^26+8x27^8
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\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\right)=0\Leftrightarrow x=2010\)
\(a,A=\frac{3}{5}.\frac{5}{4}-\frac{3}{5}.\frac{1}{4}\)
\(=\frac{3}{5}.\left(\frac{5}{4}-\frac{1}{4}\right)\)
\(=\frac{3}{5}.1=\frac{3}{5}\)
\(b,B=6,3+\left(-6,3\right)+4,9\)
\(=0+4,9=4,9\)
b, \(A=\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)
\(A=13+...+3^9.13=13\left(1+...+3^9\right)⋮13\)
`Answer:`
A) \(A=1+3+3^2+3^3+3^4+...+3^{11}\)
\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{12}\)
\(\Rightarrow3A-A=3^{12}-1\)
\(\Rightarrow A=\frac{3^{12}-1}{2}\)
B) Ta có:
\(1+3+3^2=13⋮13\)
\(3^3+3^4+3^5=3^3.\left(1+3+3^2\right)⋮13\)
...
\(3^9+3^{10}+3^{11}=3^9.\left(1+3+3^2\right)⋮13\)
\(\Rightarrow A⋮13\)