\(A=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{^{5^2}}{26\cdot31}\)
Những câu hỏi liên quan
bài 1 tính nhanh a) Afrac{2}{1cdot3}+frac{2}{3cdot5}+frac{2}{5cdot7}+...+frac{2}{99cdot101}b) Bfrac{3}{1cdot3}+frac{3}{3cdot5}+frac{3}{57}+...+frac{3}{49cdot51}c) Cfrac{5^2}{1cdot6}+frac{5^2}{6cdot11}+frac{5^2}{11cdot16}+frac{5^2}{16cdot21}+frac{5^2}{21cdot26}+frac{5^2}{26cdot31}d) Dfrac{10}{56}+frac{10}{140}+frac{10}{260}+...+frac{10}{1400}e) Efrac{3}{5cdot11}+frac{5}{11cdot21}+frac{7}{21cdot35}+frac{9}{35cdot53}f) Ffrac{2}{15}+frac{2}{35}+frac{2}{99}+frac{4}{77}giải chi tiết giúp mình nhé than...
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bài 1 tính nhanh
a) A=\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
b) B=\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{57}+...+\frac{3}{49\cdot51}\)
c) C=\(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)
d) D=\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
e) E=\(\frac{3}{5\cdot11}+\frac{5}{11\cdot21}+\frac{7}{21\cdot35}+\frac{9}{35\cdot53}\)
f) F=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{99}+\frac{4}{77}\)
giải chi tiết giúp mình nhé thank you very much
A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101
A = 2 - 2/101 = 200/101
B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51
B = 3-3/51(tự tính nhé)
C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31
C = 5(5-1/31)(tự tính)
D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)
2E nhân lên rồi giải giống trên
3F Rồi nhân 4/77 và rút gọn thì tính được
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a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0
A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
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a) A= \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)
=\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right).\frac{3}{2}\)
=\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right).\frac{3}{2}\)
= \(\left(1-\frac{1}{50}\right).\frac{3}{2}=\frac{49}{50}.\frac{3}{2}=\frac{147}{100}\)
c) \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
= \(\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right).5\)
= \(\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right).5\)
= \(\left(1-\frac{1}{31}\right).5=\frac{30}{31}.5=\frac{150}{31}\)
Mấy bài còn lại mik đang phải nháp đã. Bạn thông cảm cho mik
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Xem thêm câu trả lời
\(s=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}\)
tại sao
\(S=5\times\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}\right)\)
\(=5\times\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\right)\)
\(=5\times\left(1-\frac{1}{16}\right)\)
\(=5\times\frac{15}{16}=\frac{75}{16}\)
Vậy \(S=\frac{75}{16}\)
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=5*(1-1/6+1/6-1/11+1/11-1/16)
=5*(1-1/16)
=5-5/16
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Xem thêm câu trả lời
tính nhanh a, frac{3}{2cdot5}+frac{3}{5cdot8}+frac{3}{8cdot11}+...+frac{3}{17cdot20}b, frac{5^2}{1cdot6}+frac{5^2}{6cdot11}+frac{5^2}{11cdot16}+...+frac{5^2}{26cdot31}c, frac{1}{3}+frac{1}{9}-frac{1}{27} frac{2}{3}-frac{1}{4}+frac{5}{11} frac{5}{3}+frac{5}{9}-frac{5}{27} frac{5}{12}+1-frac{7}{11}
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tính nhanh
a, \(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\)
b, \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+...+\frac{5^2}{26\cdot31}\)
c, \(\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\) \(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}\)
\(\frac{5}{3}+\frac{5}{9}-\frac{5}{27}\) \(\frac{5}{12}+1-\frac{7}{11}\)
a,=1/2-1/5+1/5-1/8+1/8-1/11+...+1/17-1/20
=1/2-1/20=19/20
b,=5.(1-1/6+1/6-1/11+...+1/26-1/31)
=5.(1-1/31)=5.30/31 =150/31
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\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
24 tháng 3 2022 lúc 10:10
Biến đổi biểu thức sau:
\(\frac{5}{1\cdot6}=\frac{1}{?}+\frac{1}{?}\)
a) Từ đó, tính giá trị biểu thức:
\(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{2017\cdot2022}\)
b) Chứng minh \(B< A\)biết:
\(B=\frac{1}{6^2}+\frac{1}{11^2}+\frac{1}{16^2}+...+\frac{1}{2022^2}\)
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
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HT
tính nhanh :
\(B=\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+\frac{1}{11\cdot15}+\frac{1}{15\cdot19}+\frac{1}{19\cdot23}+\frac{1}{23\cdot27}+\frac{1}{27\cdot31}+\frac{1}{31\cdot35}\)
\(A=\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
Phần 1)Đầu tiên bạn nhân B với 1 phần 4 rồi tính đến đoạn gần cuối sẽ ra 1/3 - 1/35 rồi quy đòng rồi tính sẽ ra kêt quả cuối là 32/105 nha
Mình lười lắm nên chỉ help 1 phần thui nha sr
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Bài 1:a,left(x-frac{1}{2}right)cdotleft(frac{1}{2}+frac{1}{6}+frac{1}{12}+frac{1}{20}+......+frac{1}{90}right)frac{1}{3}b,frac{1}{1cdot6}+frac{1}{6cdot11}+.....+frac{1}{96cdot101}frac{1}{10cdot x}c,460+85cdot4frac{x+175}{5}+30d,left(x-5right)cdotleft(10-9frac{40}{41}right):left(1-frac{81}{82}right):left(1-frac{204}{205}right)2050
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Bài 1:
\(a,\left(x-\frac{1}{2}\right)\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}\right)=\frac{1}{3}\)
\(b,\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+.....+\frac{1}{96\cdot101}=\frac{1}{10\cdot x}\)
\(c,460+85\cdot4=\frac{x+175}{5}+30\)
\(d,\left(x-5\right)\cdot\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
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c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
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So sánh
\(\frac{1}{11\cdot11}+\frac{1}{10\cdot10}+...+\frac{1}{6\cdot6}+\frac{1}{5\cdot5}\) và \(\frac{7}{44}\)
Cách 2:
\(\frac{1}{11.11}+\frac{1}{10.10}+....+\frac{1}{5.5}
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Cho S_1-S_2+S_3-S_4+S_5frac{m}{n} với m, n nguyên tố cùng nhau. Biết:S_1frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}S_2frac{1}{2cdot3}+frac{1}{2cdot4}+frac{1}{2cdot5}+frac{1}{2cdot6}+frac{1}{3cdot4}+frac{1}{3cdot5}+frac{1}{3cdot6}+frac{1}{4cdot5}+frac{1}{4cdot6}+frac{1}{5cdot6}S_3frac{1}{2cdot3cdot4}+frac{1}{2cdot3cdot5}+frac{1}{2cdot3cdot6}+frac{1}{2cdot4cdot5}+frac{1}{2cdot4cdot6}+frac{1}{2cdot5cdot6}+frac{1}{3cdot4cdot5}+frac{1}{3cdot4cdot6}+frac{1}{3cdot5cdot6}+frac{1}{4cdot5cdot6}...
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Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)