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\(A=1+2+2^2+...+2^{2017}\)
\(\Rightarrow A=\dfrac{2^{2017+1}-1}{2-1}\)
\(\Rightarrow A=2^{2018}-1\)
mà \(B=2^{2018}\)
\(\Rightarrow A-B=2^{2018}-1-2^{2018}\)
\(\Rightarrow A-B=-1\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow A=2A-A=2^{2018}-1\)
\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)
\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
\(a,x=-2\Leftrightarrow A=\left(x-1\right)^3=\left(-2-1\right)^3=-3^3=-27\)
\(b,x=\frac{1}{2}\Rightarrow A=\left(x-1\right)^3=\left(\frac{1}{2}-1\right)^3=\left(-\frac{1}{2}\right)^3=-\frac{1^3}{2^3}=-\frac{1}{8}\)
\(\frac{4^{20}.20^{10}}{80^{10}.5^7}\)\(=\frac{4^{10}.4^{10}.20^{10}}{4^{10}.20^{10}.5^7}\)\(=\frac{4^{10}}{5^7}\)
\(\frac{9^{10}.6^3}{36^7.3^2}\)\(=\frac{3^5.3^2.3^{10}.6^3}{6^3.6^4.6^7.3^2}\)\(=\frac{3^{15}}{6^{11}}\)\(=\frac{3^{11}.3^4}{3^{11}.2^{11}}\)\(=\frac{3^4}{2^{11}}\)
Ta có: \(A=1+2+2^2+...+2^{2017}\)
\(2.A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2+2^2+2^3+...+2^{2018}-\left(1+2+2^2+...+2^{2017}\right)\)
\(A=2^{2018}-1\)
\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)