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a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)
= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)
= 90 . 7/90 + 194/90
= 630/90 + 194/90
= 824/90 = 412/45
b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35
= -2/5 + 3/7 - 4/9 - 3/5 + 13/35 + 7/35
= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)
= -1 + 3/7 - 4/9 + 4/7
= -1 + (3/7 + 4/7) - 4/9
= -1 + 1 - 4/9 = -4/9
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
ta có: \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=1-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}-\frac{1}{9}+\frac{1}{10}\)
\(=1-\frac{1}{2}+\frac{1}{10}=\frac{3}{5}\)
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
Ta có:
\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}:\frac{90}{7}\)
\(=74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)
\(=\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)
\(=\frac{7}{90}.90\)
\(=7\)
\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)
\(\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)
\(\frac{7}{90}.90\)
\(7\)