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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
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
a) Ta có:
\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)
\(A=-21\cdot26\)
\(A=-546\)
\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)
\(B=2\cdot12\cdot5\)
\(B=2\cdot60\)
\(B=120\)
Mà: \(120>-546\)
\(\Rightarrow B>A\)
Tính các tổng sau:
1, S=1-2+3_4+..+25-26
S =-1+3-5+7-...-53+55 ( có 28 số hạng )
= (-1+3)+(-5+7)+...+(-53+55) ( có 28:2=14 nhóm )
= 2+2+...+2
= 2 . 14
= 28
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
a: \(=\dfrac{7+12-6}{13}=1\)
b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)
c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)
d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
cảm ơn bn, mình đặt câu hỏi, bn thườg xuyên trả lời câu hỏi của mình. Thank you very much.
a) 3
b) 1