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\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
\(a,3\frac{1}{5}-2\frac{1}{6}+\frac{1}{5}\)
\(3\frac{1}{5}=\frac{16}{5};2\frac{1}{6}=\frac{13}{6}\)
\(MSC=30\)
\(\Rightarrow3\frac{1}{5}-2\frac{1}{6}+\frac{1}{5}\)
\(=\frac{96}{30}-\frac{65}{30}+\frac{6}{30}\)
\(=\frac{96-65+6}{30}=\frac{37}{30}\)
\(b,5\frac{12}{13}-2\frac{15}{26}+\frac{1}{26}\)
\(5\frac{12}{13}=\frac{77}{13};2\frac{15}{26}=\frac{67}{26}\)
\(MSC=26\)
\(\Rightarrow5\frac{12}{13}-2\frac{15}{26}+\frac{1}{26}\)
\(=\frac{154}{26}-\frac{134}{26}+\frac{1}{26}\)
\(=\frac{154-134+1}{26}=\frac{21}{26}\)
\(c,7\frac{1}{12}+\frac{1}{12}-6\frac{1}{4}\)
\(7\frac{1}{12}=\frac{85}{12};6\frac{1}{4}=\frac{25}{4}\)
\(MSC=12\)
\(\Rightarrow7\frac{1}{12}+\frac{1}{12}-6\frac{1}{4}\)
\(=\frac{85}{12}+\frac{1}{12}-\frac{25}{4}\)
\(=\frac{85+1-25}{12}=\frac{61}{12}\)
\(\frac{12}{7}.\left(19-15\right).\left(\frac{5}{8}-\frac{1}{4}\right)\)
=\(\frac{12}{7}.4\frac{3}{8}\)
=\(\frac{12}{7}.\frac{35}{8}\)
\(=\frac{15}{2}\)
đề cương của cậu giống hệt của tớ như 2 giọt nước
b)\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=-\frac{1}{3}\)
c)\(=\left(\frac{10}{3}+\frac{25}{10}\right):\left(\frac{19}{6}-\frac{21}{5}\right)-\frac{11}{31}\)
\(=\left(\frac{10}{3}+\frac{5}{2}\right):\left(\frac{95}{30}-\frac{126}{30}\right)-\frac{11}{31}\)
\(=\left(\frac{20}{6}+\frac{15}{6}\right):-\frac{31}{30}-\frac{11}{31}\)
\(=\frac{35}{6}:\frac{-31}{30}-\frac{11}{31}\)
\(=\frac{175}{31}-\frac{11}{31}\)
\(=-6\)
a, =\(\left(19\frac{5}{8}-15\frac{1}{4}\right):\frac{7}{12}\)
= \(\frac{35}{8}:\frac{7}{12}\)
= \(\frac{15}{2}\)
b, = \(\left(\frac{2}{5}+\frac{3}{5}\right).\frac{1}{3}\)\(-\frac{2}{3}\)
= \(1.\frac{1}{3}-\frac{2}{3}\)
= \(\frac{1}{3}-\frac{2}{3}=\frac{-1}{3}\)
c, = \(\left(\frac{10}{3}+\frac{25}{10}\right):\left(\frac{19}{6}-\frac{21}{5}\right)-\frac{11}{31}\)
= \(\frac{175}{30}:\frac{31}{30}-\frac{11}{31}\)
= \(\frac{175}{31}-\frac{11}{31}\)
= \(\frac{164}{31}\)