Tính \(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+.....+\frac{1}{96\cdot101}\)
\(s=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}\)
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 \(S=\frac{75}{16}\)
=5*(1-1/6+1/6-1/11+1/11-1/16)
=5*(1-1/16)
=5-5/16
Tính nhanh bằng cách thuận tiện nhất
a) 10,11+11,12+11,13+...+98,99+99,10
b) \(\frac{1}{1\cdot6}\cdot\frac{1}{6\cdot11}\cdot\frac{1}{11\cdot16}\cdot...\cdot\frac{1}{491\cdot496}\cdot\frac{1}{496\cdot501}\)
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
a) A = \(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot5}+...+\frac{4}{107\cdot111}\)
b) B = \(\frac{6}{15\cdot18}+\frac{6}{18\cdot21}+\frac{6}{21\cdot24}+...+\frac{6}{87\cdot90}\)
c) C = \(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{51\cdot56}\)
d) D = \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
a) A = 1/3 - 1/7 + 1/7 - 1/11 +......+1/107 - 1/111
A = 1/3 - 1/111
A = ..............Bạn tự tính nhé!
b) B = 2.(3/15.18 + 3/18.21 +........+3/87.90)
B = 2.(1/15 - 1/18 + 1/18 - 1/21 +........+1/87 - 1/90)
B = 2.(1/15 - 1/90)
B = 2.5/90
B =......Tự tính nhé!
C ; D làm tương tự nhé!
a)
A=\(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{107.111}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-.....+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{108}{333}\)
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:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
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
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-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
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
a, \(\frac{80}{1\cdot6}+\frac{80}{6\cdot11}+\frac{80}{11\cdot16}+...+\frac{80}{251\cdot256}\)
b, 1.4+2.5+3.6+4.7+...+100.103
c, tìm x:
\(\left(2x-1\right)^2\)=\(\frac{1}{4}\)
Thử trí nha
Nhanh nhất, đúng nhất mk tick cho♥
Tìm x biết \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot101}+...+\frac{1}{10\cdot101}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
tìm x biết:
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot101}+...+\frac{1}{10\cdot101}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
bài này khó quá thui tick cho mk đi mk chỉ có 1 đỉm hỏi đáp thui !