B=4.6 6.8 8.10 10.12 ... 216.2018
giải rõ cho mình nha
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Đặt A = 8.10 + 10.12 + 12.14 + ....... + 98.100
=> 6A = 8.10.12 - 8.10.12 + 10.12.14 - 10.12.14 + ...... + 98.100.102
=> 6A = 98.100.102
=> A = 98.100.102/6
=> A = 166600
c.1.2.3+2.3.4+4.5.6+5.6.7=6+24+120+210
=30+120+210
=150+210
=360
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
Ta có \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\)
\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\)
\(\Rightarrow A=\dfrac{65}{132}\)
Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\)
Vậy \(A< 1\)
\(\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
=\(3.\left(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
=\(\frac{3}{2}.\left(\frac{7}{28}-\frac{2}{28}\right)\)
=\(\frac{3}{2}.\frac{5}{28}=\frac{15}{56}\)
\(\sqrt[]{\frac{ }{ }\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^2}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
Gọi biều thức trên là A, ta có:
A=(1/2.4+1/4.6+1/6.8+1/8.10+1/10.12)x=2
2A=(2/2.4+2/4.6+2/6.8+2/8.10+2/10.12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/12)x=2
2A=5/12x=2
=>A=5/24x=1
=>x=1:5/24=24/5
=>1/2.(5/12).x=1
5/24.x=1
x=1:5/24
x=24/5
lưu ý, 1/2.5/12 là tính xong phần 1/2.4 +...+1/10.12 rùi nhé
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
Tích B tận cùng là 0
Vì bất kì số nào nhân với 10 đều tận cùng là 0
Bài này là tìm chữ số tận cùng thì rất dễ
ko đề bài yêu cầu tính ra