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Đặt \(A=\frac54+\frac58+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
=>\(A\times2=\frac52+\frac54+\frac58+\frac{5}{16}+\frac{5}{32}\)
=>\(2\times A-A=\frac52+\frac54+\frac58+\frac{5}{16}+\frac{5}{32}-\frac54-\frac58-\frac{5}{16}-\frac{5}{32}-\frac{5}{64}\)
=>\(A=\frac52-\frac{5}{64}=\frac{160}{64}-\frac{5}{64}=\frac{155}{64}\)
Ta có: \(1+\frac54+\frac58+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
\(=1+\frac{155}{64}=\frac{64+155}{64}=\frac{219}{64}\)
\(a,\left(16.23+16.77\right)-\left(5.30-5.20\right)\)
\(=16.\left(23+77\right)-5.\left(30-20\right)\)
\(=16.100-5.10\)
\(=1600-50\)
\(=1550\)
\(b,8\frac{1}{2}\div\frac{17}{5}+\frac{6}{8}\div3\frac{2}{3}\)
\(=\frac{17}{2}.\frac{5}{17}+\frac{3}{4}\div\frac{11}{3}\)
\(=\frac{17}{2}.\frac{5}{17}+\frac{3}{4}.\frac{3}{11}\)
\(=\frac{5}{2}+\frac{9}{44}\)
\(=\frac{110}{44}+\frac{9}{44}\)
\(=\frac{119}{44}\)
\(a,=16.100-5.10=1600-50=1550\)
\(b,8\frac{1}{2}:\frac{17}{5}+\frac{6}{8}:3\frac{2}{3}=\frac{17}{2}.\frac{5}{17}+\frac{6}{8}:\frac{11}{3}=\frac{5}{2}+\frac{18}{88}=\frac{220}{88}+\frac{18}{88}=\frac{238}{88}=2\frac{31}{44}\)
\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-4\right)\)
= \(\frac{6}{7}+\frac{1}{8}-\frac{-3}{4}\)
= \(\frac{55}{56}-\frac{-3}{4}\)
= \(\frac{97}{56}\)
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
mình xin nhầm đề là: A = \(1*2*3+2*4*6+4*8*12+8*16*24 \over2*3*4+4*6*8+8*12*16+16*24*32\)
ta có: 1+1/2+2+1/4+...+9+1/512
=(1+2+3+4+...+9)+(1/2+1/4+...+1/512)
=45+(1/2+1/4+...+1/512)
gọi số hạng (1/2+1/4+...+1/512) là a ta được :
a=1/2+1/4+...+1/512
2a=1+1/2+1/4+1/8+...+1/256
2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)
=1-1/512
=511/512
vậy kết quả của biểu thức đó là45+511/512
