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\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
Mình mới học lớp 5 , xin lỗi nhé, mình cũng rất muốn giúp bạn nhưng ko đc.
nếu không làm được thì thôi, mong bạn đừng nhắn lời xin lỗi ạ. Không ai như bạn đâu!
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}\right)+\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Ta có: \(A=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
Nhận xét: \(\frac{a}{x.\left(x+a\right)}=\frac{1}{x}-\frac{1}{x+a}\)
Do đó: \(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(\frac{100}{100}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\frac{99}{100}\)
\(=\frac{33}{50}\)
Vậy,\(A=\frac{33}{50}\)
\(\text{Ta có: }A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)
\(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(\Rightarrow\frac{3}{2}A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow\frac{3}{2}A=1-\frac{1}{100}\)
\(\Rightarrow\frac{3}{2}A=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}:\frac{3}{2}\)
\(A=\frac{99}{100}.\frac{2}{3}=\frac{33}{50}\)
A=2/1.4+2/4.7+2/7.10+...+2/97.100
=2/3(3/1.4+3/4.7+3/7.10+...+3/97.100)
=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
=2/3(1-1/100)=33/50
\(S=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+......+\frac{2}{97.100}\)
\(\Rightarrow S=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\\ \Rightarrow S=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\\ \Rightarrow S=\frac{2}{3}\left(1-\frac{1}{100}\right)\\ \Rightarrow S=\frac{33}{50}\)
s = 2.3 (1/1.4 +1/4.7 +..........+1/97.100):3
S = 2 :3 .(3/1.4 +3/4.7 +.............+3/97 .100)
S = 2:3 .(1/1 -1/4 +1/4 - 1/7 +1/7 -...........+1/97 -1/100 )
S = 2/3 .(1/1 -1/100)
S =2/3 .99/100 =33 /50
A= 2/1.4+2/4.7+2/7.10+...+2/97.100
= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)
= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
= 2.(1/1-1/100)
= 2.(99/100)
=99/50
\(\frac{2}{1.4}+\frac{2}{4.7}+....+\frac{2}{97.100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\left(\frac{200}{100}-\frac{2}{100}\right)=\frac{1}{3}.\frac{198}{100}=\frac{33}{50}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=\frac{297}{100}\)
\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)
\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)
\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)
\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)
\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)
=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=> \(2.\left(1-\dfrac{1}{100}\right)\)
=>\(2\).\(\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)
`Answer:`
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+2/97.100`
`=>A : 2 = 1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100`
`=>A: 2 .3 = 3/1.4 + 3/4.7 + 3/7.10 +...+3/97.100`
`=>A . 3/2 = 1/1 - 1/4 + 1/4- 1/7 +....+1/97-1/100`
`=>A. 3/2 = 1 - 1/100`
`=>A . 3/2 = 99/100`
`=>A = 99/100: 3/2`
`=>A = 99/100 . 2/3`
`=>A = 33/50`