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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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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