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\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\)
\(A=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}.\frac{7}{15}\)
\(A=\frac{7}{30}\)
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1}{2}.\frac{1009}{2020}\)
\(\Leftrightarrow A=\frac{1009}{4040}\)
Vậy : \(A=\frac{1009}{4040}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{38}-\frac{1}{40}\)
=\(\frac{1}{2}-\frac{1}{40}\)
=\(\frac{19}{40}\)
= 2 *[1/2 * 1/4 +1/4 *1/6 +1/6*1/8+...+1/38*1/40
=2*[1/2 - 1/40]
=2 * (-19/40)
= -380
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
cho mình nha!
Đặt BT trên là A
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)
\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)
Đặt A là biểu thức trên ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
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
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{200+202}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{200}-\frac{1}{202}\)
= \(\frac{1}{2}-\frac{1}{202}\)
= \(\frac{404}{202}-\frac{1}{202}\)
= \(\frac{403}{202}\)
bạn nhân 2 vào thì sẽ hiểu cách làm.