Tính
3/1x4 + 3/4x7 + 3/7x10 + ... + 3/94x97
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\(S=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{94\times97}+\frac{1}{97\times100}\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\frac{99}{100}\)
\(S=\frac{33}{100}\)
Bài này giống toán lớp 6 hơn
m = 3/(1x4) + 3/(4x7) + ... + 3/(19x22)
= (4-1)/(1x4) + (7-4)/(4x7) + ... + (22-19)/(19x22)
= 4/(1x4) - 1/(1x4) + 7/(4x7) - 4/(4x7) + ... + 22/(19x22) - 19/(19x22)
= 1 - 1/4 + 1/4 - 1/7 + ... + 1/19 - 1/22
= 1-1/22
= 21/22
=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16
=1-1/16=15/16
A=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2.\left(x+3\right)}\)
=> A=\(\frac{3}{1}-\frac{3}{4}+\frac{3}{4}+...+\frac{3}{2.x}-\frac{3}{2.\left(x+3\right)}\)
=> A =\(\frac{3}{1}-\frac{3}{2.\left(x+3\right)}\)
\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)
\(A=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3\times\left(1-\frac{1}{100}\right)\)
\(A=3\times\frac{99}{100}\)
\(A=\frac{297}{100}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
Bài làm:
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)