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Ta có:
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}.\frac{48}{49}=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+\frac{4.6}{5.5}+\frac{5.7}{6.6}+\frac{6.8}{7.7}=\frac{1.2.3.4.5.6}{2.3.4.5.6.7}.\frac{3.4.5.6.7.8}{2.3.4.5.6.7}=\frac{1}{7}.\frac{8}{2}=\frac{4}{7}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
\(A=\frac{8}{9}.\ \frac{15}{16}.\ \frac{24}{25}.\ \frac{35}{36}.\ \frac{48}{49}.\ \frac{63}{64}\)
\(A=\frac{2.4}{3^2}.\ \frac{3.5}{4^2}.\ \frac{4.6}{5^2}.\ \frac{5.7}{6^2}.\ \frac{6.8}{7^2}.\ \frac{7.9}{8^2}\)
\(A=\frac{2.3.4^2.5^2.6^2.7^2.8.9}{3^2.4^2.5^2.6^2.7^2.8^2}\)
\(A=\frac{2.9}{3.8}\)
\(A=\frac{3}{4}\)
mình biết đáp án là : \(\frac{9}{16}\)thôi,còn cách giải thì mình không chắc chắn nên không viết ra
\(\frac{3.2.4.3.5.4.6.5.7.6.8.7.9}{4.3.3.4.4.5.5.6.6.7.7.8.8}\)= \(\frac{9}{16}\)
\(a.\frac{27.45+27.55}{2+4+6+...+14+16+18}=\frac{27.100}{\frac{\left(2+18\right).9}{2}}=30\)
\(b.\frac{26.108-26.12}{32-28+24-20+16-12+8-4}=\frac{26\left(108-12\right)}{\left(32-28\right).4}=\frac{26.96}{4.4}=156\)
\(c.\frac{27.4500+135.550.2}{2+4+6+...+14+16+18}=\frac{270.450+270.550}{\frac{\left(2+8\right).9}{2}}\)
\(d.\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}=\frac{48.700-48.450}{5.5}\)\(=\frac{48\left(700-450\right)}{25}=\frac{48.250}{25}=480\)
#ĐinhBa
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
a) \(\frac{-6}{35}.\frac{-49}{60}=\frac{\left(-6\right).\left(-49\right)}{35.60}=\frac{\left(-6\right).\left(-7\right).\left(-7\right)}{\left(-5\right).\left(-7\right).\left(-6\right).\left(-10\right)}=\frac{-7}{\left(-5\right).\left(-10\right)}=\frac{-7}{50}\)
b) \(\frac{-16}{25}.\frac{-5}{24}=\frac{\left(-16\right).\left(-5\right)}{25.24}=\frac{\left(-8\right).2.\left(-5\right)}{\left(-5\right).\left(-5\right).\left(-8\right).\left(-3\right)}=\frac{2}{\left(-5\right).\left(-3\right)}=\frac{2}{15}\)
c) \(\frac{15}{-17}.\frac{5}{18}=\frac{15.5}{\left(-17\right).18}=\frac{5.3.5}{\left(-17\right).3.6}=\frac{5.5}{\left(-17\right).6}=\frac{25}{-102}\)
a) \(\frac{-6}{35}\cdot\frac{-49}{60}\)
\(=\frac{\left(-6\right)\cdot\left(-49\right)}{35\cdot60}\)
\(=\frac{294}{2100}=\frac{7}{50}\)
b)\(\frac{-16}{25}\cdot\frac{-5}{24}\)
\(=\frac{\left(-16\right)\cdot\left(-5\right)}{25\cdot24}\)
\(=\frac{80}{600}=\frac{2}{15}\)
c)\(\frac{15}{-17}\cdot\frac{5}{18}\)
\(=\frac{15\cdot5}{\left(-17\right)\cdot18}\)
\(=\frac{75}{-306}=\frac{25}{-102}\)