\(=\left(1+\dfrac{7}{3}-\dfrac{13}{4}\right):\left(1+\dfrac{43}{12}-\dfrac{9}{2}\right)\)
\(=\dfrac{12+28-39}{12}:\dfrac{12+43-54}{12}\)
\(=\dfrac{1}{1}=1\)
\(a,1\frac{13}{15}.0,75-\left(\frac{8}{15}+0,25\right).\frac{24}{47}\)
\(b,5:\left(4\frac{3}{4}-1\frac{25}{28}\right)-1\frac{3}{8}:\left(\frac{3}{8}+\frac{9}{20}\right)\)
\(c,6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(d,\left(\frac{3}{5}+0,415-\frac{3}{200}\right).2\frac{2}{3}.0,25\)
\(e,\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(g,1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right);\frac{7}{3}\)
\(\frac{10+\frac{9}{2}+\frac{8}{3}+\frac{7}{4}+ \frac{6}{5}+\frac{5}{6}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{11}}\)
\(\left(20+9\frac{1}{4}\right):2\frac{1}{4}\) \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:\frac{1}{4}\)
\(\frac{32}{15}:\left(-1\frac{1}{5}+1\frac{1}{3}\right)\) \(0,2.\frac{15}{36}-\left(\frac{2}{5}=\frac{2}{3}\right):1\frac{1}{5}\)
\(\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\) \(0,7.2\frac{2}{3}.20.0,375.\frac{5}{8}\)
Bài 1: Tìm x, biết: a) \(1\frac{1}{2}x+\frac{-4}{5}=4\) b) \(\frac{2}{3}x+\frac{-1}{2}x=\frac{-5}{12}\) c) \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)
Cho M =\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\) .Hãy chứng minh M<\(\frac{3}{16}\)
Câu 2 Chứng minh rằng :
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
\(a,\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right).\frac{15}{78}\)
\(b,\frac{-7}{21}+\left(1+\frac{1}{3}\right)\) \(c,\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(d,\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(e,\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
Bài 1:So sánh Avà B biết rằng:
A=\(\frac{10^{15}+1}{10^{16}+1};\) B=\(\frac{10^{16}+1}{10^{17}+1}\)
A=\(\frac{3}{8^3}+\frac{7}{8^4}\); B=\(\frac{7}{8^3}+\frac{3}{8^4}\)
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.......+\frac{1}{19}+\frac{1}{20};\) B=\(\frac{1}{2}\)
Bài 2:Dạng tính tổng đặc biệt:
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+......+\frac{3^2}{340}\)
\(D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^8}\)
\(E=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{99}\right)\)
Bài 3:Dạng chứng minh:
\(A=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{99}.\)Chứng minh rằng A chia hết cho 100
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\).Chứng minh rằng A>\(\frac{4}{3}\)
Bài 1: Tính(hợp lý nếu có thể) a) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\) b) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\) c) \(\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\) d) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}\)
\(C=\left(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}\right):\left(\frac{5}{12}+1-\frac{7}{11}\right)\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0\)
