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\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}\right)+\frac{2}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{2}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Mình mới học lớp 5 , xin lỗi nhé, mình cũng rất muốn giúp bạn nhưng ko đc.
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
\(\frac{2}{1.4}+\frac{2}{4.7}+....+\frac{2}{97.100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\left(\frac{200}{100}-\frac{2}{100}\right)=\frac{1}{3}.\frac{198}{100}=\frac{33}{50}\)
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
A = \(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
A = \(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)= \(\frac{2}{3}.\frac{99}{100}\)= \(\frac{33}{50}\)
a) \(16+\left(27-7\cdot6\right)-\left(94\cdot7-27\cdot99\right)\)
\(=16+27-7\cdot6-94\cdot7+27\cdot99\)
\(=16+27\left(1+99\right)-7\left(6+94\right)=16+2700-700=2016\)
b)\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+\frac{2}{7}-\frac{2}{10}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\cdot\frac{99}{50}=\frac{33}{50}\)
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=2/3(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
A=2/3(1-1/100)
A=2/3.99/100
A=33/50
mình k pit co dung k nua nghe
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
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=\frac{297}{100}\)
=2/3(3/1*4+3/4*7+...+3/97*100)
=2/3(1-1/4+1/4-1/7+...+1/97-1/100)
=2/3*99/100
=198/300
=66/100
=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
quá dễ bạn ạ