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1+\(\frac{5}{4}+\frac{5}{8}+\frac{5}{32}+\frac{5}{64}\)
= 3\(\frac{7}{64}\)
= \(\frac{199}{64}\)
Lời giải:
$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$
$2\times A-A=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64})$
$A=1-\frac{1}{64}=\frac{63}{64}$
a,234.4+234.5+234=234(4+5+1)=234.10=2340
b,135.16-135.2-135.4=135(16-4-2)=135.10=1350
c,1/2+1/4+1/8+1/16+1/32+1/64=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32=1-1/32=31/32
a) 234 x4 + 234 x 5 + 234 = 234 x ( 1+4+5)=2340
b)135 x16 -135 x2 -4 x135 =135 x (16-2-4) =1350
\(a)\frac{2}{5}+\frac{3}{10}-\frac{1}{2}\)
\(=\frac{4}{10}+\frac{3}{10}-\frac{5}{10}=\frac{1}{5}\)
\(b)\frac{8}{11}+\frac{8}{33}.\frac{3}{4}\)
\(=\frac{8}{11}+\frac{2}{11}=\frac{10}{11}\)
\(c)\frac{7}{9}.\frac{3}{14}:\frac{5}{8}\)
\(=\frac{1}{6}:\frac{5}{8}=\frac{4}{15}\)
\(d)\frac{5}{12}-\frac{7}{32}:\frac{21}{16}\)
\(=\frac{5}{12}-\frac{1}{6}=\frac{5}{12}-\frac{2}{12}=\frac{1}{4}\)
2/5 + 3/10 - 1/2
= 4/10 + 3/10 -5/10
=2/10
=1/5.
8/11 + 8/33 *3/4
=8/11 + 2/11
=10/11
7/9 *3/14 : 5/8
=1/3 *1/2 : 5/8
=1/6: 5/8
=1/3 * 4/5
=4/15
5/12 - 7/32 : 21/16
=5/12 - 1/6
=5/12 - 2/12
=1/4
Đặt \(A=1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
\(=5\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2\cdot B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow B=2\cdot B-B=1-\frac{1}{64}=\frac{63}{64}\)
\(\Rightarrow A=5\cdot\frac{63}{64}=\frac{315}{64}\)