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Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)
Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)
Chờ chút nhá :D
Lời giải:
** Sửa đề:
$A=\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+....+\frac{1}{2013}(1+2+3+...+2013)$
$A=\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+....+\frac{1}{2013}.\frac{2013.2014}{2}$
$=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2014}{2}$
$=\frac{3+4+5+...+2014}{2}$
$=\frac{1+2+3+4+5+...+2014}{2}-\frac{3}{2}$
$=\frac{2014.2015:2}{2}-\frac{3}{2}$
$=1014551$
a) \(\left(0,25+\frac{3}{4}:1,25+1\frac{1}{3}:2\right)\)\(=\left(\frac{1}{4}+\frac{3}{4}\right):\frac{5}{4}+\frac{4}{3}:2\)\(=\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\)
b) \(2^3+3\left(\frac{-3}{2}\right)^0.\left(\frac{1}{2}\right)^2.4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)\(=8+3.1.\frac{1}{4}.4+\left[4:\frac{1}{2}\right]:8\)
\(=8+3+8:8=\frac{19}{8}\)
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947
a.
\(-2^3+2^2+\left(-1\right)^{2013}=-8+4-1=-5\)
b.
\(\left(3^3\right)^2-\left[\left(-2\right)^3\right]^2-\left(-5\right)^2=27^2-\left(-8\right)^2-25=729-64-25=640\)
c.
\(2^3+3\times\left(-\frac{1}{2016}\right)^0-\left(\frac{1}{2}\right)^2\times4-\left[\left(-2\right)^2\div\frac{1}{2}\right]=8+3\times0-\frac{1}{4}\times4-\left(4\times2\right)=8+3-1-8=2\)