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Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
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Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
\(P=...\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)
\(=\frac{1}{99}-1=\frac{-98}{99}\)
\(M=...\)
\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)
\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)
\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)
\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)
a, \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.......1\frac{1}{99}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}\)
\(=\frac{\left(2.3.4......10\right)\left(2.3.4....10\right)}{\left(1.2.3....9\right)\left(3.4.5....11\right)}\)
\(=\frac{10.2}{1.11}=\frac{20}{11}\)
b, Gọi A = \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot\cdot\cdot\cdot\frac{60}{2}\),gọi B = \(1.3.5....59\)
Ta có: \(A=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}.....\frac{60}{2}\)
\(=\frac{31.32.33....60}{2^{30}}\)
\(=\frac{\left(31.32.33.....60\right)\left(1.2.3....30\right)}{2^{30}.\left(1.2.3....30\right)}\)
\(=\frac{1.2.3.....60}{\left(2.1\right)\left(2.2\right)\left(2.3\right)....\left(2.30\right)}\)
\(=\frac{1.2.3.....60}{2.4.6....60}\)
\(=\frac{\left(1.3.5...59\right)\left(2.4....60\right)}{2.4.6...60}\)
\(=1.3.5....59=B\)
Vậy A = B