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Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
toi nay mk se lay y kien cua cac ban , giup mk nha
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)
\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)
\(c.\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)
\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{3}\times\left[\left(\frac{-3}{4}+\frac{1}{4}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)
\(=\frac{7}{3}\times\frac{1}{2}\)
\(=\frac{7}{6}\)
\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)
\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)
\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)
\(=\frac{7}{8}\times\frac{44}{7}\)
\(=\frac{11}{2}\)
a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)
\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)
\(A=\dfrac{-1}{10.2}\)
\(A=\dfrac{-1}{20}\)
b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)
\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)
\(B=\dfrac{2}{3.3}\)
\(B=\dfrac{2}{9}\)
c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)
\(C=\dfrac{2}{3}+\dfrac{2}{3}\)
\(C=\dfrac{4}{3}\)
d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)
\(D=\dfrac{-57}{57}=-1\)
e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)
\(E=\dfrac{27}{182}-\dfrac{27}{182}\)
\(E=0\)
a: \(A=-5\cdot\left(-8\right)=40\)
b: \(B=-1\cdot\left(81-16\right)=-65\)
c: \(C=125\cdot\left(-7-8\right)+4=-1871\)
d: \(D=-9+8-1=-2\)