Chứng minh rằng P>3 biet P= \(\dfrac{5}{2×1}+\dfrac{4}{1×11}+\dfrac{3}{11×2}+\dfrac{1}{2×15}+\dfrac{13}{15×4}+\dfrac{15}{4×43}+\dfrac{13}{43×8}\)
Những câu hỏi liên quan
1, dfrac{-5}{7} . dfrac{2}{11} + dfrac{-5}{7} . dfrac{9}{11} + 1dfrac{5}{7}
2,-3dfrac{4}{13} . 15dfrac{3}{41} + 3dfrac{4}{13} . 2dfrac{3}{41}
3, dfrac{4}{5} . 15dfrac{1}{4} - dfrac{4}{5} . 15dfrac{1}{3} + dfrac{11}{30}
4,dfrac{4}{20} + dfrac{16}{42} + dfrac{6}{15} - dfrac{3}{5} + dfrac{2}{21} - dfrac{10}{21} + dfrac{3}{10}
Giúp mik nha. Cảm ơn
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1, \(\dfrac{-5}{7}\) . \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) . \(\dfrac{9}{11}\) + \(1\dfrac{5}{7}\)
2,\(-3\dfrac{4}{13}\) . \(15\dfrac{3}{41}\) + \(3\dfrac{4}{13}\) . \(2\dfrac{3}{41}\)
3, \(\dfrac{4}{5}\) . \(15\dfrac{1}{4}\) - \(\dfrac{4}{5}\) . \(15\dfrac{1}{3}\) + \(\dfrac{11}{30}\)
4,\(\dfrac{4}{20}\) + \(\dfrac{16}{42}\) + \(\dfrac{6}{15}\) - \(\dfrac{3}{5}\) + \(\dfrac{2}{21}\) - \(\dfrac{10}{21}\) + \(\dfrac{3}{10}\)
Giúp mik nha. Cảm ơn
19 tháng 9 2017 lúc 12:00
tinh nhanh
A=\(\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
B=\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}\)
giup minh nhe minh dang can gap
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
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tinh nhanh
A=\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}
B=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}
giup minh nhe minh dang can gap
Tính (Tính hợp lí nếu có thể)a) dfrac{-7}{12}-dfrac{3}{36}b) (4-dfrac{5}{12}):2+dfrac{5}{24}c) dfrac{8}{9}+dfrac{1}{9}.dfrac{2}{13}+dfrac{1}{9}.dfrac{11}{13}d) dfrac{3}{4}.dfrac{8}{9}.dfrac{15}{16}. ... .dfrac{9999}{10000}e) dfrac{3}{1.4}+dfrac{3}{4.7}+dfrac{3}{7.10}+...+dfrac{3}{97.100}*Lưu ý: Mong các anh chị trình bày chi tiết để em có thể hiểu bài, em xin các anh chị đừng viết mỗi kết quả xong em chả biết một cái gì ;-;
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Tính (Tính hợp lí nếu có thể)
a) \(\dfrac{-7}{12}\)-\(\dfrac{3}{36}\)
b) (4-\(\dfrac{5}{12}\)):2+\(\dfrac{5}{24}\)
c) \(\dfrac{8}{9}\)+\(\dfrac{1}{9}\).\(\dfrac{2}{13}\)+\(\dfrac{1}{9}\).\(\dfrac{11}{13}\)
d) \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\). ... .\(\dfrac{9999}{10000}\)
e) \(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
*Lưu ý: Mong các anh chị trình bày chi tiết để em có thể hiểu bài, em xin các anh chị đừng viết mỗi kết quả xong em chả biết một cái gì ;-;
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
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Tính hợp lý : \(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}\)
\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{12}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)
\(=\dfrac{1}{2}\)
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\(\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{50-\dfrac{2}{11}+\dfrac{8}{13}-\dfrac{8}{15}}=\dfrac{25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}}{2.\left(25-\dfrac{1}{11}+\dfrac{4}{13}-\dfrac{4}{15}\right)}\)
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e) \(\left(15-6\dfrac{13}{18}\right)\):\(12\dfrac{1}{27}\)-\(2\dfrac{1}{8}\):\(1\dfrac{11}{40}\)
g) (-3,2).\(\dfrac{-15}{64}\)+\(\left(0,8-2\dfrac{4}{15}\right)\):\(3\dfrac{2}{3}\)
Đề bài là:Tính các giá trị biểu thức sau ạ
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a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)
b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
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Cho A = \(\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}\) ; B = \(\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng minh rằng A= B
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
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SGK Cánh diều - Tập 2 - Trang 51
28 tháng 10 2023 lúc 0:42
Tính rồi rút gọn (theo mẫu):Mẫu: dfrac{9}{10}-dfrac{4}{10}dfrac{9-4}{10}dfrac{5}{10}dfrac{1}{2}a) dfrac{15}{8}-dfrac{13}{8} b) dfrac{7}{15}-dfrac{2}{15} c) dfrac{11}{12}-dfrac{2}{12} d) dfrac{19}{7}-dfrac{5}{7}
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Tính rồi rút gọn (theo mẫu):
| Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
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