\(\frac{7}{8}\)- \(\frac{7}{16}\)- \(\frac{11}{32}\)= ?
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A=\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
A=[ \(\frac{1}{3}-\frac{1}{3}\)] + [ \(-\frac{3}{5}+\frac{3}{5}\)] + [ \(-\frac{5}{7}+\frac{5}{7}\)] + [ \(-\frac{7}{9}+\frac{7}{9}\)] + [ \(-\frac{9}{11}+\frac{9}{11}\)] \(-\frac{11}{13}\)
Các bạn tự làm tiếp nhé!Sorry
=7(1/2 +1/4 +1/8+1/16+1/32+1/64+1/128)
Xét 1/2+1/4+1/8+1/16+1/32+1/64+1/128
= 1+1/2+1/4+1/8+1/16+1/32+1/64 - 1/2-1/4-1/8-1/16-1/32-1/64-1/128
=1-1/128
=127/128
Vậy tổng ban đầu là:7 * 127/128=889/128
Ta có: \(A=\frac{\frac{3}{11}+1-\frac{3}{7}}{3+\frac{9}{11}-\frac{9}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
\(=\frac{3\left(\frac{1}{11}+\frac{1}{3}-\frac{1}{7}\right)}{9\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{7}\right)}-\frac{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}\)
\(=\frac{3}{9}-\frac{2}{7}=\frac{1}{3}-\frac{2}{7}=\frac{7}{21}-\frac{6}{21}=\frac{1}{21}\)
Vậy \(A=\frac{1}{21}\)
\(\frac{7}{8}-\frac{7}{16}-\frac{11}{32}=\frac{28}{32}-\frac{14}{32}-\frac{11}{32}=\frac{3}{32}\)
\(\frac{3}{32}\)
ban nhe
tk nhe
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