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\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
~~ Bố thí cái li.ke ~~
Ta có :
A= 1+ 1/2 + 1/3 +1/4 + ...+ 1/63 + 1/64
=1 + ( 1/2 + 1/3 + 1/4 ) + ( 1/5 +1/6 + ..+1/8 ) + ( 1/9 + 1/10 + ..+ 1/16 ) + ( 1/17 + 1/18 + ...+ 1/32 ) + ( 1/33 + 1/34 + ...+1/63 + 1/64 )
=> A > 1 + ( 1/2 + 1/4.2 ) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32
A > 1 + 1/2 + 1/2 + 1/2 +1/2
=>A > 4
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
\(1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-.......-\frac{2}{61.63}-\frac{2}{63.65}\)
=\(-1.\left(\frac{2}{3.5}+\frac{2}{5.7}+......\frac{2}{63.65}\right)+1\)
=\(-1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{63}-\frac{1}{65}\right)+1\)
=\(-1.\left(\frac{1}{3}-\frac{1}{65}\right)+1\)
=\(-1.\frac{62}{195}+1\)
=\(\frac{-62}{195}+\frac{195}{195}\)
=\(\frac{133}{195}\)
Hok tốt nhé bn
\(D=\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+\frac{15}{2^4}+\frac{31}{2^5}+\frac{63}{2^6}=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}\)\(=\frac{1}{2}+\left(1-\frac{1}{4}+1-\frac{1}{8}+1-\frac{1}{16}+1-\frac{1}{32}+1-\frac{1}{64}\right)=\frac{1}{2}+5-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)đến đấy thôi
\(\Rightarrow\left(\frac{\frac{31}{36}}{\frac{11}{9}:4-\frac{3}{4}}:31\right).x=\frac{-1}{16}\Rightarrow\left(\frac{\frac{31}{36}}{\frac{-4}{9}}:31\right)x=-\frac{1}{16}\Rightarrow\left(-\frac{31}{16}:31\right)x=-\frac{1}{16}\Rightarrow-\frac{1}{16}x=-\frac{1}{16}\Rightarrow x=1\)
