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Mình không nghĩ TH làm được bài này đâu nên mình làm cách THCS nha, dù sao đây cũng là toán lớp 6 mà!!!
Gọi tổng là A ta có :
\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{6561}\)
\(A=\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\)
\(3A=3.\left(\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\right)\)
\(3A=\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+...+\frac{2}{3^9}\)
\(3A-A=\left(\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+...+\frac{2}{3^9}\right)-\left(\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+...+\frac{2}{3^8}\right)\)
\(2A=\frac{2}{3^9}-\frac{2}{3^1}\)
\(A=\frac{\frac{2}{3^9}-\frac{2}{3^1}}{2}\)
Vậy,...
Nếu sai mong bạn thông cảm nha!!!
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)
\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)
\(=\dfrac{231}{54}:\dfrac{7}{12}\)
\(=\dfrac{198}{27}\)
\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)
\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)
\(=\dfrac{0,8.0,64}{0,6}\)
\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
Sửa đề: A=1/3+1/9+1/27+...+1/6561
=1/3+1/3^2+1/3^3+...+1/3^8
=>3A=1+1/3+...+1/3^7
=>3A-A=1-1/3^8
=>\(2A=\dfrac{3^8-1}{3^8}\)
=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)
Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)
\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)
\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)
\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)
\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)
\(2S=\dfrac{4132}{6561}\)
\(S=\dfrac{2066}{6561}\)