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a, \(\frac{5}{7}+\frac{8}{7}=\frac{13}{7}\)
b,\(\frac{9}{8}-\frac{3}{8}=\frac{6}{8}\)
c,\(\frac{6}{1}:\frac{6}{5}=\frac{6}{1}\cdot\frac{5}{6}=\frac{30}{6}=5\)
~~~Hok tốt ~~~
a,\(\frac{13}{7}\)
b,\(\frac{6}{8}\)= \(\frac{3}{4}\)
c,\(\frac{9}{7}\)
d,5
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)
a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)
\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)
\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)
\(=\frac{5}{11}\times\frac{22}{25}\)
\(=\frac{2}{5}\)
b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
\(\frac{18}{20}\)=\(\frac{9}{10}\)vì mấy phép tính trên đổi sai hết đơn vị.
\(a,\)\(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(b,\)\(\sqrt{4}+\sqrt[3]{8}+\frac{2}{3}\)
\(=2+2+\frac{2}{3}\)
\(=4+\frac{2}{3}\)
\(=\frac{14}{3}\)
\(9-\frac{13}{4}+\frac{4}{3}\div2=\frac{36}{4}-\frac{13}{4}+\frac{4}{3}\times\frac{1}{2}\)
\(=\frac{23}{4}+\frac{2}{3}\)
\(=\frac{77}{12}\)
đây nè:A=0. tui chắc chắn lun.