Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)
a, 1,5+1-0,75/2,5+5\3-1,25
=15\10+1-75\100/25\10+5\3-125\100
=7\4/35/12
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+-\frac{37}{100}+\frac{1}{8}+-\frac{128}{100}+-\frac{250}{100}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{37}{12}\right)-\left(\frac{37}{100}+\frac{128}{100}+\frac{250}{100}\right)\)
\(=\left(\frac{114}{24}+\frac{3}{24}+\frac{74}{24}\right)-\frac{415}{100}\)
\(=\frac{191}{24}-\frac{415}{100}\)
\(=\frac{457}{120}\)
Tham khảo nha !!!
\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+\frac{-37}{100}+\frac{1}{8}+\frac{-32}{25}+\frac{-5}{2}+\frac{37}{12}\)
\(=\left(\frac{19}{4}+\frac{1}{8}+\frac{-5}{2}\right)+\left(\frac{-37}{100}+\frac{-32}{25}\right)+\frac{37}{12}\)
\(=\frac{19}{8}+\frac{-33}{20}+\frac{37}{12}\)
\(=\frac{29}{40}+\frac{37}{12}\)
\(=\frac{457}{120}\)
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
\(=\frac{16}{3}x\frac{30}{7}=\frac{480}{21}=\frac{160}{7}\)
\(\frac{33}{5}:\frac{21}{4}=\frac{33}{5}x\frac{4}{21}=\frac{132}{105}=\frac{44}{35}\)
a/ \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+........+\frac{99}{100!}\)
\(\Leftrightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+......+\frac{100-1}{100!}\)
\(\Leftrightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+.....+\frac{100}{100!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{99!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{100!}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}\)
=(1975/1976+2010/2011+1963/1968)x(4/12-3/12-1/12)
=(1975/1976+2010/2011+1963/1968)x0
=0