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A) 150-[102-(14-11)2 .20210
=150-[102-32.20210 ]
=150-[100-9.1]
=150-91
= 59
A = \(\dfrac{2^{2021}+1}{2^{2021}}\) = \(\dfrac{2^{2021}}{2^{2021}}\) + \(\dfrac{1}{2^{2021}}\) = 1 + \(\dfrac{1}{2^{2021}}\)
B = \(\dfrac{2^{2021}+2}{2^{2021}+1}\) = \(\dfrac{2^{2021}+1+1}{2^{2021}+1}\) = \(\dfrac{2^{2021}+1}{2^{2021}+1}\) +\(\dfrac{1}{2^{2021}+1}\) = 1 + \(\dfrac{1}{2^{2021}+1}\)
Vì \(\dfrac{1}{2^{2021}}\) > \(\dfrac{1}{2^{2021}+1}\) nên 1 + \(\dfrac{1}{2^{2021}}\) > 1 + \(\dfrac{1}{2^{2021}+1}\)
Vậy A > B
\(3^2\left(x+4\right)-5^2=5\cdot2^2\\ 9\left(x+4\right)-25=20\\ 9\left(x+4\right)=20+25\\ 9\left(x+4\right)=45\\ x+4=45:9\\ x+4=5\\ x=5-4\\ x=1\)
\(3^2.5+2^3.10-3^4:3\)
\(=5\left(3^2+2^3.2\right)-3^{4-1}\)
\(=5\left(9+16\right)-3^3\)
\(=5.25-27\)
\(=125-27=98\)
\(3^2\times5+2^3\times10-3^4:3\\ =9\times5+8\times10-27\\ =45+80-27\\ =98.\)
\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}=\frac{30}{30}=1\)
\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}=\frac{3}{4}-\frac{2}{4}+\frac{2}{5}=\frac{1}{4}+\frac{2}{5}=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}\)
Chúc bạn học tốt!
\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}\)
\(=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}\)
\(=\frac{18+5+7}{30}\)
\(=\frac{30}{30}\)
\(=1\)
\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}\)
\(\frac{15}{20}-\frac{10}{20}+\frac{8}{20}\)
\(=\frac{15-10+8}{20}\)
\(=\frac{13}{20}\)
\(1+4+16+...+4^{2021}\)
Đặt biểu thức trên là \(A\), ta có:
\(A=1+4+16+...+4^{2021}\)
\(A=1+4+4^{2}+...+4^{2021}\)
\(4A=4+4^{2}+4^{3}+...+4^{2022}\)
\(4A-A=(4+4^{2}+4^{3}+...+4^{2022})-(1+4+4^{2}+...+4^{2021})\)
\(3A=4^{2022}-1\)
\(A=\dfrac{4^{2022}-1}{3}\)