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\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
A = \(\frac{\frac{\frac{5+3.3-1.12}{12}}{3.6-5+2.2}+}{6}+\frac{16\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}=\frac{\frac{\frac{5+9-12}{12}}{18-5+4}}{6}+\frac{16}{17}=\frac{2}{12}.\frac{6}{17}+\frac{16}{17}=\frac{1}{17}.\frac{6}{17}=1\)
A=\(\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{18}{6}-\frac{5}{6}+\frac{4}{6}}+\frac{16.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}\)
=\(\frac{\frac{1}{6}}{\frac{17}{6}}+\frac{16}{17}\)
=\(\frac{1}{17}+\frac{16}{17}\)
=1
Cac ban giup mik voi, mik k cho! ( 10 k cho 3 nguoi dau tien tra loi cau hoi cua mik)
m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))
= \(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)
= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)
= 0+1+\(\frac{7}{17}\)
= \(\frac{24}{17}\)
n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))
= \(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)
= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1
= 1+(-1)+1
= 0+1
= 1
o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)
= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= \(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{32}{5}\)
= \(\frac{18}{5}\)
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