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\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Trả lời
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{4.7}+...+\frac{3}{97.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Ta thấy :
\(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
\(.........\)
\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)
đáp án = \(\frac{297}{100}\)
đúng không?
kết bạn với mh nha
\(B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)
A:3=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\)
A:3=\(\frac{1}{1}-\frac{1}{100}\)
A:3=\(\frac{99}{100}\)
A=\(\frac{99}{100}.3\)
A=\(\frac{297}{100}\)
\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(A:3=\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A:3=\frac{1}{1}-\frac{1}{100}\)
\(A:3=\frac{99}{100}\)
\(A=\frac{99}{100}.3\)
\(A=\frac{297}{100}\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}\left(\frac{1}{7}-\frac{1}{10}\right)+..........+\frac{1}{3}\left(\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+........+\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{100}\right)\)
=\(\frac{1}{3}x\frac{6}{25}\)=\(\frac{2}{25}\)
vậy biểu thức trên có giá trị bằng\(\frac{2}{25}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=\frac{297}{100}\)
Vậy \(A=\frac{297}{100}\)
wow hay thiệt mình làm giống kết quả nhưng khác cách giải thanks bạn nhìu nha
Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)
1/4-1/7 = 3/28 = 3.(1/4.7)
A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)
A = 3.(1-1/100)
A = 3.(99/100)
A = 297/100
\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}\)
\(A=\frac{33}{100}\)
tớ ko biết
S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)