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a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
\(=\frac{3}{200}:\frac{3}{5}+\frac{3}{2}\left(\frac{4}{25}-\frac{2}{5}\right)-\frac{1}{25}.\left(\frac{7}{4}:\frac{7}{5}-\frac{5}{2}\right)\)
\(=\frac{3.5}{200.3}+\frac{3}{2}\left(\frac{4}{25}-\frac{2.5}{25}\right)-\frac{1}{25}\left(\frac{7.5}{4.7}-\frac{5}{2}\right)\)
\(=\frac{1}{40}+\frac{3}{2}\left(\frac{-6}{25}\right)-\frac{1}{25}\left(\frac{5}{4}-\frac{10}{4}\right)\)
\(=\frac{1}{40}-\frac{9}{25}+\frac{1}{20}\)
\(=\frac{1.5}{40.5}-\frac{9.8}{25.8}+\frac{1.10}{20.10}\)
\(=\frac{5-72+10}{200}=\frac{-57}{200}\)
\(=\frac{108}{63}+\frac{105}{63}=\frac{108+105}{63}=\frac{213}{63}\)
\(\frac{12}{7}\)-\(\frac{-15}{9}\) = \(\frac{12}{7}\)+ \(\frac{15}{9}\) = \(\frac{108}{63}\)+ \(\frac{105}{63}\)= \(\frac{108+109}{63}\)= \(\frac{217}{63}\)= \(\frac{31}{9}\)
nguyễn thị thu huyền nhầm rồi 217/63 chứ
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
=>\(A=2A-A=2+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
\(A=2+\frac{1}{2^{98}}\)
Vậy: \(A=2+\frac{1}{2^{98}}\)
Gọi \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2B-B=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{100}}\)
\(\Rightarrow A=2\)
Vậy A = 2
ta có: C = 1/32 + 1/34 + 1/36 +...+ 1/3100 => 9C = 1 + 1/32 +1/34 +...+1/398
=> 9C - C = (1 + 1/32 + 1/34 +...+1/398 ) - (1/32 +1/34 + 1/36 +...+ 1/3100)
=> 8C = 1 - 1/3100 => C = (1 - 1/3100 ) / 8
đúng ko nhỉ
Gọi \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(\Rightarrow2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
=> 2A - A = \(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow A=1+\frac{1}{64}=\frac{65}{64}\)
=\(\frac{3\left(\frac{1}{1}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{2}{4}+\frac{2}{6}+\frac{2}{8}}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)
=\(\frac{3}{5}+\frac{2\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)=\(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
Ta có: \(A=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+......+\frac{5}{64.69}\right)\)
\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+......+\frac{1}{64}-\frac{1}{69}\right)\)
\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{69}\right)=\frac{1}{5}\times\frac{65}{276}\)
\(\Rightarrow A=\frac{13}{276}\)
Vậy \(A=\frac{13}{276}\)
\(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{64.69}\)
\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
\(5A=\frac{1}{4}-\frac{1}{69}\)
\(A=\frac{65}{276}:5\)
\(A=\frac{13}{276}\)