1. tính
A=\(\frac{1}{1\cdot6}\)+\(\frac{1}{6\cdot11}\)+\(\frac{1}{11\cdot16}\)+\(\frac{1}{16\cdot21}\)+..........+\(\frac{1}{496\cdot501}\)
làm ơn giúp mik
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Ta có
5y = 5/(1.6) + 5/(6.11) + ... + 5/(496.501)
= 1-1/6+1/6-1/11+...+1/496-1/501
= 1-1/501
= 500/501
=>y = 100/501
Gọi A=1/1x6+1/6x11+1/11x16+...+1/96x101
A=1/1x6+1/6x11+1/11x16+...+1/96x101
5A=5/1x6+5/6x11+5/11x16+...+5/96x101
5A=1-1/6+1/6-1/11+1/11-1/16+...+1/96-1/101
5A=1-1/101
5A=100/101
A=100/101:5
A=20/101.
Nếu đúng thì kết bạn với mình nhé.Mình học lớp 6C của trường THCS An Khê.Tên Đỗ Đường Hùng.
= 1/5 . (5/1.6 + 5/6.11 + ...... + 5/96.101)
= 1/5 . (1-1/6+1/6-1/11+.....+1/96-1/101)
= 1/5.(1-1/101)
= 1/5 . 100/101
= 20/101
Tk mk nha
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) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
@@@@@@@@@@
HT
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-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
\(S=5\times\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}\right)\)
\(=5\times\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\right)\)
\(=5\times\left(1-\frac{1}{16}\right)\)
\(=5\times\frac{15}{16}=\frac{75}{16}\)
Vậy \(S=\frac{75}{16}\)
\(A=\frac{15}{1.6}+\frac{15}{6.11}+\frac{15}{11.16}+...+\frac{15}{2011.2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+..+\frac{5}{2011.2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{2011}-\frac{1}{2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=1-\frac{1}{2016}\)
\(\Rightarrow\)\(\frac{1}{3}A=\frac{2015}{2016}\)
\(\Rightarrow\)\(A=\frac{2015}{672}\) (1)
Mà \(3=\frac{2016}{672}\) (2)
Từ (1) và (2) suy ra A < 3
Trả lời
a)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(\Leftrightarrow x^2=\frac{11}{4}\cdot\frac{16}{11}\)
\(\Leftrightarrow x^2=\frac{16}{4}\)
\(\Leftrightarrow x^2=\left(\frac{4}{2}\right)^2\)
\(\Leftrightarrow x=\frac{4}{2}\)
Vậy x=\(\frac{4}{2}\)
b) (bạn thiếu nhóm \(\frac{1}{10\cdot13}\))
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\)
\(\Rightarrow3A=3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\right)\)
\(\Rightarrow3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\)
\(\Rightarrow3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}\)
\(\Rightarrow3A=1-\frac{1}{19}\Leftrightarrow3A=\frac{18}{19}\)
\(\Rightarrow A=\frac{18}{19}:3\Leftrightarrow A=\frac{6}{19}\)
\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{491.496}+\frac{1}{496.501}\)
\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{491.496}+\frac{5}{496.501}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{491}-\frac{1}{496}+\frac{1}{496}-\frac{1}{501}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{501}\right)=\frac{1}{5}.\frac{500}{501}=\frac{100}{501}\)