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\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)
Đặt \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\frac{102}{103}\)
\(\Rightarrow B=\frac{68}{103}\)
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\cdot\frac{102}{103}\)
\(A=\frac{68}{103}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{33}{34}=\dfrac{11}{17}\)
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)
Dấu \(.\)là dấu nhân
Ta có :
\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)
\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)
\(\Rightarrow E=\frac{68}{103}\)
Vậy \(E=\frac{68}{103}\)
~ Ủng hộ nhé
\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)
\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)
Gọi tổng trong ngoặc là F
\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)
\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)
\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)
\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)
Vậy......
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\)
\(=1-\dfrac{1}{40}\)
\(=\dfrac{39}{40}\)
1/(1×4) + 1/(4×7) + ... + 1/(34×37) + 1/(37×40)
= 1/3 × (1 - 1/4 + 1/4 - 1/7 + ... + 1/34 - 1/37 + 1/37 - 1/40)
= 1/3 × (1 - 1/40)
= 1/3 × 39/40
= 13/40
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)
\(=1-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=1-\dfrac{1}{40}\)
\(=\dfrac{39}{40}\)
Vậy giá trị cần tìm là: \(\dfrac{39}{40}\)
\(3xA=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+...+\dfrac{37-34}{34x37}+\dfrac{40-37}{37x40}=\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{17}{ }+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}=\)
\(=1-\dfrac{1}{40}=\dfrac{39}{40}\Rightarrow A=\dfrac{39}{40}:3=\dfrac{13}{40}\)
3/(1×4)+3/(4×7)+3/(7×10)+3/(10×13)+3/(13×16)
=1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16
=1-1/16
=15/16
31 x 434 x 737 x 10310 x 13 = 1.3289876e+12
mik phải dùng máy tính chứ có sịp nhân mới trả lời đc
nhỉ ?????
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
\(\frac{2}{1\times4}+\frac{2}{4\times7}+\frac{2}{7\times10}+...+\frac{2}{37\times40}\)
\(=\frac{2}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{37\times40}\right)\)
\(=\frac{2}{3}\times\left(\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+...+\frac{40-37}{37\times40}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{40}\right)=\frac{13}{20}\)